{"id":1496,"date":"2018-08-19T03:12:11","date_gmt":"2018-08-19T03:12:11","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/?post_type=chapter&#038;p=1496"},"modified":"2024-11-22T18:31:35","modified_gmt":"2024-11-22T18:31:35","slug":"9-3-solution-stoichiometry","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/9-3-solution-stoichiometry\/","title":{"raw":"9.3  Solution Stoichiometry","rendered":"9.3  Solution Stoichiometry"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>earning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Perform stoichiometric calculations involving solution molarity<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"ball-ch11_s04_p13\" class=\"para editable block\">As we have seen in lab, many reactions such as single or double displacement reactions are carried out in aqueous medium (i.e. in water).\u00a0 Because these reactions occur in aqueous solution, we can use the concept of molarity to directly calculate the number of moles of reactants or products that will be formed, and hence their amounts (i.e. volume of solutions or mass of precipitates).<\/p>\r\n<p class=\"para editable block\">More complex stoichiometry problems using balanced chemical reactions can also use concentrations as conversion factors.<\/p>\r\n<p class=\"para editable block\">For example, suppose the following equation represents a chemical reaction:<\/p>\r\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large{\\text{2 Ag}}{\\text{NO}}_{3}\\text{(}aq\\text{)}{\\text{ + CaCl}}_{2}\\text{(}aq\\text{)}\\rightarrow {\\text{2 AgCl}}\\text{(}s\\text{)}\\text{ + Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}\\text{(}aq\\text{)}[\/latex]<\/span><\/span><\/p>\r\n<p id=\"ball-ch11_s04_p14\" class=\"para editable block\">If we wanted to know what volume of 0.555 M CaCl<sub class=\"subscript\">2<\/sub> would react with 1.25 mol of AgNO<sub class=\"subscript\">3<\/sub>, we first use the balanced chemical equation to determine the number of moles of CaCl<sub class=\"subscript\">2<\/sub> that would react and then use molarity to convert to liters of solution:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large1.25\\cancel{\\text{ mol Ag}{\\text{NO}_{3}}}\\times \\frac{1\\cancel{\\text{ mol Ca}{\\text{Cl}}_{2}}}{2\\cancel{\\text{ mol Ag}{\\text{NO}_{3}}}}\\times \\frac{1\\cancel{\\text{ L solution}}}{0.555\\cancel{\\text{ mol Ca}{\\text{Cl}_{2}}}}=\\text{1.13}\\text{ L Ca}{\\text{Cl}_{2}}{\\text{ solution}}[\/latex]<\/p>\r\n<p id=\"ball-ch11_s04_p15\" class=\"para editable block\">This can be extended by starting with the mass (grams) of one reactant, instead of moles of a reactant.<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 1: Solution Stoichiometry--Mass to Volume Conversion<strong>\r\n<\/strong><\/h3>\r\nWhat volume of 0.0995 M Al(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub> will react with 3.66 g of Ag according to the following chemical equation?\r\n<p style=\"text-align: center;\">[latex]\\text{3 Ag}\\text{(}s\\text{)}\\text{ + Al}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{3}\\text{(}aq\\text{)}\\rightarrow {\\text{3 Ag}}\\text{NO}_{3}\\text{(}aq\\text{)}\\text{ + Al}\\text{(}s\\text{)}[\/latex]<\/p>\r\n[reveal-answer q=\"1595961\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"1595961\"]\r\n<p id=\"ball-ch11_s04_p17\" class=\"para\">Here, we first must convert the mass of Ag to moles before using a mole to mole ratio from the balanced chemical equation and then the definition of molarity as a conversion factor:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\text{3.66}\\cancel{\\text{ g Ag}}\\times \\frac{1\\cancel{\\text{ mol Ag}}}{107.97\\cancel{\\text{ g Ag}}}\\times \\frac{1\\cancel{\\text{ mol Al}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{3}}}{3\\cancel{\\text{ mol Ag}}}\\times \\frac{1\\cancel{\\text{ L solution}}}{0.0995\\cancel{\\text{ mol Al}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{3}}}=\\text{0.114}\\text{ L Al}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{3}{\\text{ solution}}[\/latex]<\/p>\r\n<p class=\"para\">[\/hidden-answer]<\/p>\r\n\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nWhat volume of 0.512 M NaOH will react with 17.9 g of H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub>(s) according to the following chemical equation?\r\n<p style=\"text-align: center;\">[latex]\\text{H}_{2}{\\text{C}}_{2}{\\text{O}}_{4}\\text{(}s\\text{)}{\\text{ + 2 NaOH}}\\text{(}aq\\text{)}\\rightarrow {\\text{Na}_{2}}{\\text{C}}_{2}{\\text{O}}_{4}\\text{(}aq\\text{)}\\text{ + 2 H}_{2}\\text{O}\\text{(}l\\text{)}[\/latex]<\/p>\r\n[reveal-answer q=\"4883021\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"4883021\"] 0.777 L of NaOH solution[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"ball-ch11_s04_p21\" class=\"para editable block\">We can extend our skills even further by recognizing that we can relate quantities of one solution to quantities of another solution. Knowing the volume and concentration of a solution containing one reactant, we can determine how much of another solution of another reactant will be needed using the balanced chemical equation.<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 2: Solution Stoichiometry--Volume to Volume Conversion<strong>\r\n<\/strong><\/h3>\r\n<p id=\"ball-ch11_s04_p22\" class=\"para\">A student takes a precisely measured sample, called an <em class=\"emphasis\">aliquot<\/em>, of 10.00 mL of a solution of FeCl<sub class=\"subscript\">3<\/sub>. The student carefully adds 0.1074 M Na<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub> until all the Fe<sup class=\"superscript\">3+<\/sup>(aq) has precipitated as Fe<sub class=\"subscript\">2<\/sub>(C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">3<\/sub>(s). Using a precisely measured tube called a burette, the student finds that 9.04 mL of the Na<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub> solution was added to completely precipitate the Fe<sup class=\"superscript\">3+<\/sup>(aq). What was the concentration of the FeCl<sub class=\"subscript\">3<\/sub> in the original solution? (A precisely measured experiment like this, which is meant to determine the amount of a substance in a sample, is called a <em class=\"emphasis\">titration<\/em>.) The balanced chemical equation is as follows:<\/p>\r\n<p style=\"text-align: center;\"><span class=\"informalequation\"><span class=\"mathphrase\">[latex]\\text{2 FeCl}_{3}\\text{(}aq\\text{)}{\\text{ + 3 Na}_{2}}\\text{C}_{2}\\text{O}_{4}\\text{(}aq\\text{)}\\rightarrow {\\text{ Fe}_{2}}{\\text{(}{\\text{C}}_{2}\\text{O}_{4}\\text{)}}_{3}\\text{(}s\\text{)}\\text{ + 6 NaCl}\\text{(}aq\\text{)}[\/latex]<\/span><\/span><\/p>\r\n[reveal-answer q=\"1595962\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"1595962\"]\r\n<p style=\"text-align: center;\"><span class=\"informalequation\">[latex]\\text{9.04}\\cancel{{\\text{ mL Na}_{2}}\\text{C}_{2}\\text{O}_{4}}\\times\\frac{\\text{1}\\cancel{\\text{ L}}}{\\text{1000}\\cancel{\\text{ mL}}}\\times\\frac{{\\text{0.1074}\\cancel{{\\text{ mol Na}_{2}}\\text{C}_{2}\\text{O}_{4}}}}{\\text{1}\\cancel{\\text{ L Na}_{2}\\text{C}_{2}\\text{O}_{4}}}\\times\\frac{\\text{2}{\\text{ mol Fe}\\text{Cl}_{3}}}{\\text{3}\\cancel{\\text{ mol Na}_{2}\\text{C}_{2}\\text{O}_{4}}}\\times\\frac{\\text{1}}{\\text{0.01000 L soln}}=\\text{0.0647 M}\\text{ FeCl}_{3}[\/latex]<\/span><\/p>\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nA student titrates 25.00 mL of H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub> with 0.0987 M KOH. She uses 54.06 mL to complete the chemical reaction. What is the concentration of H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub>?\r\n<p style=\"text-align: center;\">[latex]\\text{H}_{3}{\\text{PO}}_{4}\\text{(}aq\\text{)}{\\text{ + 3 KOH}}\\text{(}aq\\text{)}\\rightarrow {\\text{K}_{3}}{\\text{PO}}_{4}\\text{(}aq\\text{)}\\text{ + 3 H}_{2}\\text{O}\\text{(}l\\text{)}[\/latex]<\/p>\r\n[reveal-answer q=\"4883022\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"4883022\"] 0.0711 M[\/hidden-answer]\r\n\r\n[caption id=\"attachment_3251\" align=\"alignnone\" width=\"600\"]<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/4269191327_7f6f150338_b.jpg\"><img class=\"wp-image-3251 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14214053\/4269191327_7f6f150338_b-e1412018548367-1.jpg\" alt=\"When a student performs a titration, a measured amount of one solution is added to another reactant. \u201cChemistry titration lab\u201d by Kentucky Country Day is licensed under the Creative Commons Attribution-NonCommercial 2.0 Generic.\" width=\"600\" height=\"398\" \/><\/a> When a student performs a titration, a measured amount of one solution is added to another reactant. \u201cChemistry titration lab\u201d by Kentucky Country Day is licensed under the Creative Commons Attribution-NonCommercial 2.0 Generic.[\/caption]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 3: Solution Stoichiometry &amp; the IDeal GAs law<strong>\r\n<\/strong><\/h3>\r\n<p id=\"ball-ch11_s04_p22\" class=\"para\">How many liters of carbon dioxide gas can form at STP when 125 mL of a 2.25 M HCl solution reacts with excess lithium carbonate?<\/p>\r\n<p style=\"text-align: center;\"><span class=\"informalequation\"><span class=\"mathphrase\">[latex]\\text{2 HCl}\\text{(}aq\\text{)}{\\text{ + Li}_{2}}\\text{C}\\text{O}_{3}\\text{(}aq\\text{)}\\rightarrow {\\text{C}}\\text{O}_{2}\\text{(}g\\text{)}\\text{ + 2 LiCl}\\text{(}aq\\text{)}{\\text{ + H}_{2}}\\text{O}\\text{(}l\\text{)}[\/latex]<\/span><\/span><\/p>\r\n[reveal-answer q=\"15959692\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"15959692\"]\r\n\r\nFirst determine the moles of carbon dioxide:\r\n<p style=\"text-align: center;\">[latex]\\text{125}\\cancel{{\\text{ mL HCl}}}\\times\\frac{\\text{1}\\cancel{\\text{ L}}}{\\text{1000}\\cancel{\\text{ mL}}}\\times\\frac{{\\text{2.25}\\cancel{{\\text{ mol HCl}}}}}{\\text{1}\\cancel{\\text{ L HCl}}}\\times\\frac{\\text{1}{\\text{ mol C}\\text{O}_{2}}}{\\text{2}\\cancel{\\text{ mol HCl}}}=\\text{0.140625 mol}\\text{ CO}_{2}[\/latex]<\/p>\r\nNext, use the ideal gas law to calculate the volume of carbon dioxide:\r\n<p style=\"text-align: center;\">[latex]{V}=\\frac{nRT}{P}=\\frac{\\left(0.140625\\cancel{\\text{ mol CO}_{2}}\\right){(0.0821\\frac{\\text{L}\\cdot \\cancel{\\text{atm}}}{\\cancel{\\text{K}} \\cdot \\cancel{\\text{mol}}})}\\left(273.15\\cancel{\\text{ K}}\\right)}{(1.00\\cancel{\\text{ atm}})}=3.15\\text{ L}\\text{ CO}_{2}[\/latex]<\/p>\r\nAlternatively, since the reaction is at STP, we know 1 mol of gas equals 22.41 L of gas (known as molar volume).\r\n<p style=\"text-align: center;\">[latex]\\text{125}\\cancel{{\\text{ mL HCl}}}\\times\\frac{\\text{1}\\cancel{\\text{ L}}}{\\text{1000}\\cancel{\\text{ mL}}}\\times\\frac{{\\text{2.25}\\cancel{{\\text{ mol HCl}}}}}{\\text{1}\\cancel{\\text{ L HCl}}}\\times\\frac{\\text{1}\\cancel{{\\text{ mol C}\\text{O}_{2}}}}{\\text{2}\\cancel{\\text{ mol HCl}}}\\times\\frac{\\text{22.41}{\\text{ L C}\\text{O}_{2}}}{\\text{1}\\cancel{\\text{ mol C}\\text{O}_{2}}}=\\text{3.15 L}\\text{ CO}_{2}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">[\/hidden-answer]<\/p>\r\n\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nIf 355 mL of hydrogen gas is collected at 25 <sup>o<\/sup>C at a total pressure of 740 mmHg from the reaction of excess zinc and a 3.0 M HCl solution, how many mL of the HCl solution was required?\r\n<p style=\"text-align: center;\">[latex]\\text{Zn}\\text{(}s\\text{)}{\\text{ + 2 HCl}}\\text{(}aq\\text{)}\\rightarrow {\\text{Zn}}{\\text{Cl}}_{2}\\text{(}aq\\text{)}\\text{ + H}_{2}\\text{(}g\\text{)}[\/latex]<\/p>\r\n[reveal-answer q=\"488302219\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"488302219\"]\u00a0 9.4 mL HCl[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\nMolarity can be used as a conversion factor in combination with reaction stoichiometry.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>End of Module Problems<\/h3>\r\n1.\u00a0 Magnesium reacts with hydrochloric acid to produce hydrogen gas and magnesium chloride. How many liters of a 0.750 M HCl solution will react with 12.25 g of Mg?\r\n<p style=\"text-align: center;\">[latex]\\text{Mg}\\text{(}s\\text{)}{\\text{ + 2 HCl}}\\text{(}aq\\text{)}\\rightarrow {\\text{H}_{2}}\\text{(}g\\text{)}\\text{ + Mg}{\\text{Cl}}_{2}\\text{(}aq\\text{)}[\/latex]<\/p>\r\n2.\u00a0 Consider the following reaction:\r\n<p style=\"text-align: center;\">[latex]\\text{Pb}\\text{(}\\text{N}\\text{O}_{3}\\text{)}_{2}\\text{(}aq\\text{)}{\\text{ + 2 NaI}}\\text{(}aq\\text{)}\\rightarrow {\\text{PbI}_{2}}\\text{(}s\\text{)}\\text{ + 2 NaN}{\\text{O}}_{3}\\text{(}aq\\text{)}[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 How many grams of lead(II) iodide will be formed from 25.0 mL of a 2.00 M sodium iodide solution?<\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 How many milliliters of a 1.25 M lead(II) nitrate solution will react with 25.0 mL of a 1.50 M sodium iodide solution?<\/p>\r\n<p style=\"padding-left: 30px;\">c.\u00a0 What is the molarity of a 20.0 mL solution of sodium iodide that reacts completely with 60.0 mL of a 0.750 M lead(II) nitrate solution?<\/p>\r\n3.\u00a0 Consider the following reaction:\r\n<p style=\"text-align: center;\">[latex]\\text{Al}_{2}\\text{(}\\text{S}\\text{O}_{4}\\text{)}_{3}\\text{(}aq\\text{)}{\\text{ + 6 KOH}}\\text{(}aq\\text{)}\\rightarrow {\\text{2 Al}}\\text{(}\\text{OH}\\text{)}_{3}\\text{(}s\\text{)}\\text{ + 3 K}_{2}{\\text{SO}}_{3}\\text{(}aq\\text{)}[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 How many grams of aluminum hydroxide will be formed from 55.0 mL of a 1.50 M potassium hydroxide solution?<\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 How many milliliters of a 0.250 M aluminum sulfate solution will react with 10.0 mL of a 3.00 M potassium hydroxide solution?<\/p>\r\n<p style=\"padding-left: 30px;\">c.\u00a0 What is the molarity of a 40.0 mL solution of potassium hydroxide that reacts completely with 20.0 mL of a 0.500 M aluminum sulfate solution?<\/p>\r\n4. Copper(II) oxide reacts with hydrochloric acid to produce copper(II) chloride and water. How many liters of a 4.50 M HCl solution will react with 33.0 g of copper(II) oxide?\r\n<p style=\"text-align: center;\">[latex]\\text{CuO}\\text{(}s\\text{)}{\\text{ + 2 HCl}}\\text{(}aq\\text{)}\\rightarrow {\\text{CuCl}_{2}}\\text{(}aq\\text{)}\\text{ + H}_{2}\\text{O}\\text{(}l\\text{)}[\/latex]<\/p>\r\n5. What volume, in liters, of NO<sub>2<\/sub> gas at 750 mmHg and 25 <sup>o<\/sup>C will be required to produce 0.150 L of a 0.500 M HNO<sub>3<\/sub> solution given the reaction below?\r\n<p style=\"text-align: center;\">[latex]\\text{3 NO}_{2}(g){\\text{ + H}_{2}}\\text{O}(l)\\rightarrow {\\text{2 HNO}_{3}}(aq)\\text{ + NO}(g)[\/latex]<\/p>\r\n6. The reaction of 34.4 mL of a HCl solution reacts with excess zinc to produce 0.720 L of hydrogen gas at 740 torr and 24\u00a0<sup>o<\/sup>C.\u00a0 What will be the molarity of the original HCl solution?\r\n<p style=\"text-align: center;\">[latex]\\text{Zn}\\text{(}s\\text{)}{\\text{ + 2 HCl}}\\text{(}aq\\text{)}\\rightarrow {\\text{Zn}}{\\text{Cl}}_{2}\\text{(}aq\\text{)}\\text{ + H}_{2}\\text{(}g\\text{)}[\/latex]<\/p>\r\n[reveal-answer q=\"132180\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"132180\"]\r\n\r\n1. 1.34 L HCl solution\r\n<p style=\"text-align: center;\">[latex]\\text{12.25}\\cancel{\\text{ g Mg}}\\times \\frac{1\\cancel{\\text{ mol Mg}}}{24.31\\cancel{\\text{ g Mg}}}\\times \\frac{2\\cancel{\\text{ mol HCl}}}{1\\cancel{\\text{ mol Mg}}}\\times \\frac{1\\cancel{\\text{ L solution}}}{0.750\\cancel{\\text{ mol HCl}}}=\\text{1.34}\\text{ L HCl}{\\text{ solution}}[\/latex]<\/p>\r\n2.\r\n\r\n(a). 11.5 g PbI<sub>2<\/sub>\r\n<p style=\"text-align: center;\">[latex]\\text{25.0}\\cancel{\\text{ mL NaI}}\\times\\frac{\\text{1}\\cancel{\\text{ L}}}{\\text{1000}\\cancel{\\text{ mL}}}\\times\\frac{{\\text{2.00}\\cancel{{\\text{ mol NaI}}}}}{\\text{1}\\cancel{\\text{ L NaI}}}\\times\\frac{\\text{1}{\\cancel{\\text{ mol Pb}\\text{I}_{2}}}}{\\text{2}\\cancel{\\text{ mol NaI}}}\\times\\frac{\\text{461.01}\\text{ g Pb}\\text{I}_{2}}{\\text{1}\\cancel{\\text{ mol Pb}\\text{I}_{2}}}=\\text{11.5}\\text{ g Pb}\\text{I}_{2}[\/latex]<\/p>\r\n(b).\u00a0 15.0 mL Pb(NO<sub>3<\/sub>)<sub>2<\/sub>\r\n<p style=\"text-align: center;\">[latex]\\text{25.0}\\cancel{{\\text{ mL NaI}}}\\times\\frac{{\\text{1.50}\\cancel{{\\text{ mol NaI}}}}}{\\text{1000}\\cancel{\\text{ mL NaI}}}\\times\\frac{\\text{1}{\\cancel{\\text{ mol Pb}\\text{(}\\text{NO}_{3}\\text{)}_{3}}}}{\\text{2}\\cancel{\\text{ mol NaI}}}\\times\\frac{\\text{1000}\\text{ mL Pb}\\text{(}\\text{NO}_{3}\\text{)}_{3}}{\\text{1.25}\\cancel{\\text{ mol Pb}\\text{(}\\text{NO}_{3}\\text{)}_{3}}}=\\text{15.0}\\text{ mL Pb}\\text{(}\\text{NO}_{3}\\text{)}_{3}[\/latex]<\/p>\r\n(c). 4.5 M NaI\r\n<p style=\"text-align: center;\">[latex]\\text{60.0 }\\cancel{{\\text{mL Pb}\\text{(}\\text{NO}_{3}\\text{)}_{3}}}\\times\\frac{{\\text{0.750 }\\cancel{{\\text{mol Pb}\\text{(}\\text{NO}_{3}\\text{)}_{3}}}}}{\\text{1000 }\\cancel{{\\text{mL Pb}\\text{(}\\text{NO}_{3}\\text{)}_{3}}}}\\times\\frac{\\text{2 }{\\cancel{\\text{mol NaI}}}}{\\text{1 }\\cancel{{\\text{mol Pb}\\text{(}\\text{NO}_{3}\\text{)}_{3}}}}\\times\\frac{\\text{1}}{\\text{0.0200}\\cancel{\\text{ L NaI}}}=\\text{4.5}\\text{ M NaI}[\/latex]<\/p>\r\n3.\r\n\r\n(a). 2.15 g Al(OH)<sub>3<\/sub>\r\n<p style=\"text-align: center;\">[latex]\\text{55.0 }\\cancel{\\text{mL KOH}}\\times\\frac{{\\text{1.50 }\\cancel{{\\text{mol KOH}}}}}{\\text{1000 }\\cancel{\\text{mL KOH}}}\\times\\frac{\\text{2 }{\\cancel{\\text{mol Al}\\text{(OH)}_{3}}}}{\\text{6 }\\cancel{\\text{mol KOH}}}\\times\\frac{\\text{78.00 }\\text{g Al}\\text{(OH)}_{3}}{\\text{1 }\\cancel{\\text{mol Al}\\text{(OH)}_{3}}}=\\text{2.15 }\\text{g}\\text{ Al}\\text{(OH)}_{3}[\/latex]<\/p>\r\n(b). 20.0 mL Al<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3<\/sub>\r\n<p style=\"text-align: center;\">[latex]\\text{10.0 }\\cancel{\\text{mL KOH}}\\times\\frac{{\\text{3.00 }\\cancel{{\\text{mol KOH}}}}}{\\text{1000 }\\cancel{\\text{mL KOH}}}\\times\\frac{\\text{1 }{\\cancel{\\text{mol Al}_{2}\\text{(SO}_{4}\\text{)}_{3}}}}{\\text{6 }\\cancel{\\text{mol KOH}}}\\times\\frac{\\text{1000 }\\text{mL Al}_{2}\\text{(SO}_{4}\\text{)}_{3}}{\\text{0.250 }\\cancel{\\text{mol Al}_{2}\\text{(SO}_{4}\\text{)}_{3}}}=\\text{20.0 }\\text{mL Al}_{2}\\text{(SO}_{4}\\text{)}_{3}[\/latex]<\/p>\r\n(c). 1.50 M KOH\r\n<p style=\"text-align: center;\">[latex]\\text{20.0 }\\cancel{\\text{mL Al}_{2}\\text{(SO}_{4}\\text{)}_{3}}\\times\\frac{{\\text{0.500 }\\cancel{\\text{mol Al}_{2}\\text{(SO}_{4}\\text{)}_{3}}}}{\\text{1000 }\\cancel{\\text{mL Al}_{2}\\text{(SO}_{4}\\text{)}_{3}}}\\times\\frac{\\text{6 }{\\cancel{\\text{mol KOH}}}}{\\text{1 }\\cancel{\\text{mol Al}_{2}\\text{(SO}_{4}\\text{)}_{3}}}\\times\\frac{\\text{1 }}{\\text{0.0400 }\\text{L Al}_{2}\\text{(SO}_{4}\\text{)}_{3}}=\\text{1.50 }\\text{M KOH}[\/latex]<\/p>\r\n4. 0.184 L HCl solution\r\n<p style=\"text-align: center;\">[latex]\\text{33.0 }\\cancel{\\text{g CuO}}\\times \\frac{\\text{1 }\\cancel{\\text{mol CuO}}}{\\text{79.55 }\\cancel{\\text{g CuO}}}\\times \\frac{\\text{2 }\\cancel{\\text{mol HCl}}}{\\text{1 }\\cancel{\\text{mol CuO}}}\\times \\frac{\\text{1 }\\cancel{\\text{L solution}}}{\\text{4.50 }\\cancel{\\text{mol HCl}}}=\\text{0.184 }\\text{L HCl}{\\text{ solution}}[\/latex]<\/p>\r\n5. 2.80 L NO<sub>2<\/sub>\r\n<p style=\"text-align: center;\">[latex]\\text{0.150 }\\cancel{{\\text{L HNO}_{3}}}\\times\\frac{{\\text{0.500 }\\cancel{{\\text{mol HNO}_{3}}}}}{\\text{1 }\\cancel{\\text{L HNO}_{3}}}\\times\\frac{\\text{3 }{\\text{mol N}\\text{O}_{2}}}{\\text{2 }\\cancel{\\text{mol HCl}}}=\\text{0.113 mol}\\text{ NO}_{2}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{V}=\\frac{nRT}{P}=\\frac{\\left(\\text{0.113 }\\cancel{\\text{mol NO}_{2}}\\right){(0.0821\\large\\frac{\\text{L}\\cdot \\cancel{\\text{atm}}}{\\cancel{\\text{K}} \\cdot \\cancel{\\text{mol}}})}\\left(\\text{298 }\\cancel{\\text{K}}\\right)}{(\\text{750 }\\cancel{\\text{mmHg}}\\times\\large\\frac{\\text{1 }\\cancel{\\text{atm}}}{\\text{760 }\\cancel{\\text{mmHg}}})}=\\text{2.80 }\\text{L}\\text{ NO}_{2}[\/latex]<\/p>\r\n6. 1.67 M HCl\r\n<p style=\"text-align: center;\">[latex]{n}=\\frac{PV}{RT}=\\frac{(\\text{740 }\\cancel{\\text{torr}}\\times{}\\frac{\\text{1 }\\cancel{\\text{atm}}}{\\text{760 }\\cancel{\\text{torr}}})(\\text{0.720 }\\cancel{\\text{L}}\\text{ H}_{2})}{(0.0821\\large\\frac{\\cancel{\\text{L}}\\cdot \\cancel{\\text{atm}}}{\\cancel{\\text{K}} \\cdot \\text{mol}})(\\text{297 }\\cancel{\\text{K}})}=\\text{0.0288}\\text{ mol}\\text{ H}_{2}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\text{0.0288 }\\cancel{{\\text{mol H}_{2}}}\\times\\frac{\\text{2 }{\\text{mol HCl}}}{\\text{1 }\\cancel{\\text{mol H}_{2}}}\\times\\frac{1}{\\text{0.0344 }\\text{L solution}}=\\text{1.67 M}\\text{ HCl}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>earning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Perform stoichiometric calculations involving solution molarity<\/li>\n<\/ul>\n<\/div>\n<p id=\"ball-ch11_s04_p13\" class=\"para editable block\">As we have seen in lab, many reactions such as single or double displacement reactions are carried out in aqueous medium (i.e. in water).\u00a0 Because these reactions occur in aqueous solution, we can use the concept of molarity to directly calculate the number of moles of reactants or products that will be formed, and hence their amounts (i.e. volume of solutions or mass of precipitates).<\/p>\n<p class=\"para editable block\">More complex stoichiometry problems using balanced chemical reactions can also use concentrations as conversion factors.<\/p>\n<p class=\"para editable block\">For example, suppose the following equation represents a chemical reaction:<\/p>\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large{\\text{2 Ag}}{\\text{NO}}_{3}\\text{(}aq\\text{)}{\\text{ + CaCl}}_{2}\\text{(}aq\\text{)}\\rightarrow {\\text{2 AgCl}}\\text{(}s\\text{)}\\text{ + Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}\\text{(}aq\\text{)}[\/latex]<\/span><\/span><\/p>\n<p id=\"ball-ch11_s04_p14\" class=\"para editable block\">If we wanted to know what volume of 0.555 M CaCl<sub class=\"subscript\">2<\/sub> would react with 1.25 mol of AgNO<sub class=\"subscript\">3<\/sub>, we first use the balanced chemical equation to determine the number of moles of CaCl<sub class=\"subscript\">2<\/sub> that would react and then use molarity to convert to liters of solution:<\/p>\n<p style=\"text-align: center;\">[latex]\\large1.25\\cancel{\\text{ mol Ag}{\\text{NO}_{3}}}\\times \\frac{1\\cancel{\\text{ mol Ca}{\\text{Cl}}_{2}}}{2\\cancel{\\text{ mol Ag}{\\text{NO}_{3}}}}\\times \\frac{1\\cancel{\\text{ L solution}}}{0.555\\cancel{\\text{ mol Ca}{\\text{Cl}_{2}}}}=\\text{1.13}\\text{ L Ca}{\\text{Cl}_{2}}{\\text{ solution}}[\/latex]<\/p>\n<p id=\"ball-ch11_s04_p15\" class=\"para editable block\">This can be extended by starting with the mass (grams) of one reactant, instead of moles of a reactant.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1: Solution Stoichiometry&#8211;Mass to Volume Conversion<strong><br \/>\n<\/strong><\/h3>\n<p>What volume of 0.0995 M Al(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub> will react with 3.66 g of Ag according to the following chemical equation?<\/p>\n<p style=\"text-align: center;\">[latex]\\text{3 Ag}\\text{(}s\\text{)}\\text{ + Al}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{3}\\text{(}aq\\text{)}\\rightarrow {\\text{3 Ag}}\\text{NO}_{3}\\text{(}aq\\text{)}\\text{ + Al}\\text{(}s\\text{)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q1595961\">Show Answer<\/span><\/p>\n<div id=\"q1595961\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"ball-ch11_s04_p17\" class=\"para\">Here, we first must convert the mass of Ag to moles before using a mole to mole ratio from the balanced chemical equation and then the definition of molarity as a conversion factor:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{3.66}\\cancel{\\text{ g Ag}}\\times \\frac{1\\cancel{\\text{ mol Ag}}}{107.97\\cancel{\\text{ g Ag}}}\\times \\frac{1\\cancel{\\text{ mol Al}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{3}}}{3\\cancel{\\text{ mol Ag}}}\\times \\frac{1\\cancel{\\text{ L solution}}}{0.0995\\cancel{\\text{ mol Al}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{3}}}=\\text{0.114}\\text{ L Al}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{3}{\\text{ solution}}[\/latex]<\/p>\n<p class=\"para\"><\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>What volume of 0.512 M NaOH will react with 17.9 g of H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub>(s) according to the following chemical equation?<\/p>\n<p style=\"text-align: center;\">[latex]\\text{H}_{2}{\\text{C}}_{2}{\\text{O}}_{4}\\text{(}s\\text{)}{\\text{ + 2 NaOH}}\\text{(}aq\\text{)}\\rightarrow {\\text{Na}_{2}}{\\text{C}}_{2}{\\text{O}}_{4}\\text{(}aq\\text{)}\\text{ + 2 H}_{2}\\text{O}\\text{(}l\\text{)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q4883021\">Show Answer<\/span><\/p>\n<div id=\"q4883021\" class=\"hidden-answer\" style=\"display: none\"> 0.777 L of NaOH solution<\/div>\n<\/div>\n<\/div>\n<p id=\"ball-ch11_s04_p21\" class=\"para editable block\">We can extend our skills even further by recognizing that we can relate quantities of one solution to quantities of another solution. Knowing the volume and concentration of a solution containing one reactant, we can determine how much of another solution of another reactant will be needed using the balanced chemical equation.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2: Solution Stoichiometry&#8211;Volume to Volume Conversion<strong><br \/>\n<\/strong><\/h3>\n<p id=\"ball-ch11_s04_p22\" class=\"para\">A student takes a precisely measured sample, called an <em class=\"emphasis\">aliquot<\/em>, of 10.00 mL of a solution of FeCl<sub class=\"subscript\">3<\/sub>. The student carefully adds 0.1074 M Na<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub> until all the Fe<sup class=\"superscript\">3+<\/sup>(aq) has precipitated as Fe<sub class=\"subscript\">2<\/sub>(C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">3<\/sub>(s). Using a precisely measured tube called a burette, the student finds that 9.04 mL of the Na<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub> solution was added to completely precipitate the Fe<sup class=\"superscript\">3+<\/sup>(aq). What was the concentration of the FeCl<sub class=\"subscript\">3<\/sub> in the original solution? (A precisely measured experiment like this, which is meant to determine the amount of a substance in a sample, is called a <em class=\"emphasis\">titration<\/em>.) The balanced chemical equation is as follows:<\/p>\n<p style=\"text-align: center;\"><span class=\"informalequation\"><span class=\"mathphrase\">[latex]\\text{2 FeCl}_{3}\\text{(}aq\\text{)}{\\text{ + 3 Na}_{2}}\\text{C}_{2}\\text{O}_{4}\\text{(}aq\\text{)}\\rightarrow {\\text{ Fe}_{2}}{\\text{(}{\\text{C}}_{2}\\text{O}_{4}\\text{)}}_{3}\\text{(}s\\text{)}\\text{ + 6 NaCl}\\text{(}aq\\text{)}[\/latex]<\/span><\/span><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q1595962\">Show Answer<\/span><\/p>\n<div id=\"q1595962\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\"><span class=\"informalequation\">[latex]\\text{9.04}\\cancel{{\\text{ mL Na}_{2}}\\text{C}_{2}\\text{O}_{4}}\\times\\frac{\\text{1}\\cancel{\\text{ L}}}{\\text{1000}\\cancel{\\text{ mL}}}\\times\\frac{{\\text{0.1074}\\cancel{{\\text{ mol Na}_{2}}\\text{C}_{2}\\text{O}_{4}}}}{\\text{1}\\cancel{\\text{ L Na}_{2}\\text{C}_{2}\\text{O}_{4}}}\\times\\frac{\\text{2}{\\text{ mol Fe}\\text{Cl}_{3}}}{\\text{3}\\cancel{\\text{ mol Na}_{2}\\text{C}_{2}\\text{O}_{4}}}\\times\\frac{\\text{1}}{\\text{0.01000 L soln}}=\\text{0.0647 M}\\text{ FeCl}_{3}[\/latex]<\/span><\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>A student titrates 25.00 mL of H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub> with 0.0987 M KOH. She uses 54.06 mL to complete the chemical reaction. What is the concentration of H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub>?<\/p>\n<p style=\"text-align: center;\">[latex]\\text{H}_{3}{\\text{PO}}_{4}\\text{(}aq\\text{)}{\\text{ + 3 KOH}}\\text{(}aq\\text{)}\\rightarrow {\\text{K}_{3}}{\\text{PO}}_{4}\\text{(}aq\\text{)}\\text{ + 3 H}_{2}\\text{O}\\text{(}l\\text{)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q4883022\">Show Answer<\/span><\/p>\n<div id=\"q4883022\" class=\"hidden-answer\" style=\"display: none\"> 0.0711 M<\/div>\n<\/div>\n<div id=\"attachment_3251\" style=\"width: 610px\" class=\"wp-caption alignnone\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/4269191327_7f6f150338_b.jpg\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3251\" class=\"wp-image-3251 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14214053\/4269191327_7f6f150338_b-e1412018548367-1.jpg\" alt=\"When a student performs a titration, a measured amount of one solution is added to another reactant. \u201cChemistry titration lab\u201d by Kentucky Country Day is licensed under the Creative Commons Attribution-NonCommercial 2.0 Generic.\" width=\"600\" height=\"398\" \/><\/a><\/p>\n<p id=\"caption-attachment-3251\" class=\"wp-caption-text\">When a student performs a titration, a measured amount of one solution is added to another reactant. \u201cChemistry titration lab\u201d by Kentucky Country Day is licensed under the Creative Commons Attribution-NonCommercial 2.0 Generic.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 3: Solution Stoichiometry &amp; the IDeal GAs law<strong><br \/>\n<\/strong><\/h3>\n<p id=\"ball-ch11_s04_p22\" class=\"para\">How many liters of carbon dioxide gas can form at STP when 125 mL of a 2.25 M HCl solution reacts with excess lithium carbonate?<\/p>\n<p style=\"text-align: center;\"><span class=\"informalequation\"><span class=\"mathphrase\">[latex]\\text{2 HCl}\\text{(}aq\\text{)}{\\text{ + Li}_{2}}\\text{C}\\text{O}_{3}\\text{(}aq\\text{)}\\rightarrow {\\text{C}}\\text{O}_{2}\\text{(}g\\text{)}\\text{ + 2 LiCl}\\text{(}aq\\text{)}{\\text{ + H}_{2}}\\text{O}\\text{(}l\\text{)}[\/latex]<\/span><\/span><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q15959692\">Show Answer<\/span><\/p>\n<div id=\"q15959692\" class=\"hidden-answer\" style=\"display: none\">\n<p>First determine the moles of carbon dioxide:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{125}\\cancel{{\\text{ mL HCl}}}\\times\\frac{\\text{1}\\cancel{\\text{ L}}}{\\text{1000}\\cancel{\\text{ mL}}}\\times\\frac{{\\text{2.25}\\cancel{{\\text{ mol HCl}}}}}{\\text{1}\\cancel{\\text{ L HCl}}}\\times\\frac{\\text{1}{\\text{ mol C}\\text{O}_{2}}}{\\text{2}\\cancel{\\text{ mol HCl}}}=\\text{0.140625 mol}\\text{ CO}_{2}[\/latex]<\/p>\n<p>Next, use the ideal gas law to calculate the volume of carbon dioxide:<\/p>\n<p style=\"text-align: center;\">[latex]{V}=\\frac{nRT}{P}=\\frac{\\left(0.140625\\cancel{\\text{ mol CO}_{2}}\\right){(0.0821\\frac{\\text{L}\\cdot \\cancel{\\text{atm}}}{\\cancel{\\text{K}} \\cdot \\cancel{\\text{mol}}})}\\left(273.15\\cancel{\\text{ K}}\\right)}{(1.00\\cancel{\\text{ atm}})}=3.15\\text{ L}\\text{ CO}_{2}[\/latex]<\/p>\n<p>Alternatively, since the reaction is at STP, we know 1 mol of gas equals 22.41 L of gas (known as molar volume).<\/p>\n<p style=\"text-align: center;\">[latex]\\text{125}\\cancel{{\\text{ mL HCl}}}\\times\\frac{\\text{1}\\cancel{\\text{ L}}}{\\text{1000}\\cancel{\\text{ mL}}}\\times\\frac{{\\text{2.25}\\cancel{{\\text{ mol HCl}}}}}{\\text{1}\\cancel{\\text{ L HCl}}}\\times\\frac{\\text{1}\\cancel{{\\text{ mol C}\\text{O}_{2}}}}{\\text{2}\\cancel{\\text{ mol HCl}}}\\times\\frac{\\text{22.41}{\\text{ L C}\\text{O}_{2}}}{\\text{1}\\cancel{\\text{ mol C}\\text{O}_{2}}}=\\text{3.15 L}\\text{ CO}_{2}[\/latex]<\/p>\n<p style=\"text-align: left;\"><\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>If 355 mL of hydrogen gas is collected at 25 <sup>o<\/sup>C at a total pressure of 740 mmHg from the reaction of excess zinc and a 3.0 M HCl solution, how many mL of the HCl solution was required?<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Zn}\\text{(}s\\text{)}{\\text{ + 2 HCl}}\\text{(}aq\\text{)}\\rightarrow {\\text{Zn}}{\\text{Cl}}_{2}\\text{(}aq\\text{)}\\text{ + H}_{2}\\text{(}g\\text{)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q488302219\">Show Answer<\/span><\/p>\n<div id=\"q488302219\" class=\"hidden-answer\" style=\"display: none\">\u00a0 9.4 mL HCl<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p>Molarity can be used as a conversion factor in combination with reaction stoichiometry.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>End of Module Problems<\/h3>\n<p>1.\u00a0 Magnesium reacts with hydrochloric acid to produce hydrogen gas and magnesium chloride. How many liters of a 0.750 M HCl solution will react with 12.25 g of Mg?<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Mg}\\text{(}s\\text{)}{\\text{ + 2 HCl}}\\text{(}aq\\text{)}\\rightarrow {\\text{H}_{2}}\\text{(}g\\text{)}\\text{ + Mg}{\\text{Cl}}_{2}\\text{(}aq\\text{)}[\/latex]<\/p>\n<p>2.\u00a0 Consider the following reaction:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Pb}\\text{(}\\text{N}\\text{O}_{3}\\text{)}_{2}\\text{(}aq\\text{)}{\\text{ + 2 NaI}}\\text{(}aq\\text{)}\\rightarrow {\\text{PbI}_{2}}\\text{(}s\\text{)}\\text{ + 2 NaN}{\\text{O}}_{3}\\text{(}aq\\text{)}[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 How many grams of lead(II) iodide will be formed from 25.0 mL of a 2.00 M sodium iodide solution?<\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 How many milliliters of a 1.25 M lead(II) nitrate solution will react with 25.0 mL of a 1.50 M sodium iodide solution?<\/p>\n<p style=\"padding-left: 30px;\">c.\u00a0 What is the molarity of a 20.0 mL solution of sodium iodide that reacts completely with 60.0 mL of a 0.750 M lead(II) nitrate solution?<\/p>\n<p>3.\u00a0 Consider the following reaction:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Al}_{2}\\text{(}\\text{S}\\text{O}_{4}\\text{)}_{3}\\text{(}aq\\text{)}{\\text{ + 6 KOH}}\\text{(}aq\\text{)}\\rightarrow {\\text{2 Al}}\\text{(}\\text{OH}\\text{)}_{3}\\text{(}s\\text{)}\\text{ + 3 K}_{2}{\\text{SO}}_{3}\\text{(}aq\\text{)}[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 How many grams of aluminum hydroxide will be formed from 55.0 mL of a 1.50 M potassium hydroxide solution?<\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 How many milliliters of a 0.250 M aluminum sulfate solution will react with 10.0 mL of a 3.00 M potassium hydroxide solution?<\/p>\n<p style=\"padding-left: 30px;\">c.\u00a0 What is the molarity of a 40.0 mL solution of potassium hydroxide that reacts completely with 20.0 mL of a 0.500 M aluminum sulfate solution?<\/p>\n<p>4. Copper(II) oxide reacts with hydrochloric acid to produce copper(II) chloride and water. How many liters of a 4.50 M HCl solution will react with 33.0 g of copper(II) oxide?<\/p>\n<p style=\"text-align: center;\">[latex]\\text{CuO}\\text{(}s\\text{)}{\\text{ + 2 HCl}}\\text{(}aq\\text{)}\\rightarrow {\\text{CuCl}_{2}}\\text{(}aq\\text{)}\\text{ + H}_{2}\\text{O}\\text{(}l\\text{)}[\/latex]<\/p>\n<p>5. What volume, in liters, of NO<sub>2<\/sub> gas at 750 mmHg and 25 <sup>o<\/sup>C will be required to produce 0.150 L of a 0.500 M HNO<sub>3<\/sub> solution given the reaction below?<\/p>\n<p style=\"text-align: center;\">[latex]\\text{3 NO}_{2}(g){\\text{ + H}_{2}}\\text{O}(l)\\rightarrow {\\text{2 HNO}_{3}}(aq)\\text{ + NO}(g)[\/latex]<\/p>\n<p>6. The reaction of 34.4 mL of a HCl solution reacts with excess zinc to produce 0.720 L of hydrogen gas at 740 torr and 24\u00a0<sup>o<\/sup>C.\u00a0 What will be the molarity of the original HCl solution?<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Zn}\\text{(}s\\text{)}{\\text{ + 2 HCl}}\\text{(}aq\\text{)}\\rightarrow {\\text{Zn}}{\\text{Cl}}_{2}\\text{(}aq\\text{)}\\text{ + H}_{2}\\text{(}g\\text{)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q132180\">Show Selected Answers<\/span><\/p>\n<div id=\"q132180\" class=\"hidden-answer\" style=\"display: none\">\n<p>1. 1.34 L HCl solution<\/p>\n<p style=\"text-align: center;\">[latex]\\text{12.25}\\cancel{\\text{ g Mg}}\\times \\frac{1\\cancel{\\text{ mol Mg}}}{24.31\\cancel{\\text{ g Mg}}}\\times \\frac{2\\cancel{\\text{ mol HCl}}}{1\\cancel{\\text{ mol Mg}}}\\times \\frac{1\\cancel{\\text{ L solution}}}{0.750\\cancel{\\text{ mol HCl}}}=\\text{1.34}\\text{ L HCl}{\\text{ solution}}[\/latex]<\/p>\n<p>2.<\/p>\n<p>(a). 11.5 g PbI<sub>2<\/sub><\/p>\n<p style=\"text-align: center;\">[latex]\\text{25.0}\\cancel{\\text{ mL NaI}}\\times\\frac{\\text{1}\\cancel{\\text{ L}}}{\\text{1000}\\cancel{\\text{ mL}}}\\times\\frac{{\\text{2.00}\\cancel{{\\text{ mol NaI}}}}}{\\text{1}\\cancel{\\text{ L NaI}}}\\times\\frac{\\text{1}{\\cancel{\\text{ mol Pb}\\text{I}_{2}}}}{\\text{2}\\cancel{\\text{ mol NaI}}}\\times\\frac{\\text{461.01}\\text{ g Pb}\\text{I}_{2}}{\\text{1}\\cancel{\\text{ mol Pb}\\text{I}_{2}}}=\\text{11.5}\\text{ g Pb}\\text{I}_{2}[\/latex]<\/p>\n<p>(b).\u00a0 15.0 mL Pb(NO<sub>3<\/sub>)<sub>2<\/sub><\/p>\n<p style=\"text-align: center;\">[latex]\\text{25.0}\\cancel{{\\text{ mL NaI}}}\\times\\frac{{\\text{1.50}\\cancel{{\\text{ mol NaI}}}}}{\\text{1000}\\cancel{\\text{ mL NaI}}}\\times\\frac{\\text{1}{\\cancel{\\text{ mol Pb}\\text{(}\\text{NO}_{3}\\text{)}_{3}}}}{\\text{2}\\cancel{\\text{ mol NaI}}}\\times\\frac{\\text{1000}\\text{ mL Pb}\\text{(}\\text{NO}_{3}\\text{)}_{3}}{\\text{1.25}\\cancel{\\text{ mol Pb}\\text{(}\\text{NO}_{3}\\text{)}_{3}}}=\\text{15.0}\\text{ mL Pb}\\text{(}\\text{NO}_{3}\\text{)}_{3}[\/latex]<\/p>\n<p>(c). 4.5 M NaI<\/p>\n<p style=\"text-align: center;\">[latex]\\text{60.0 }\\cancel{{\\text{mL Pb}\\text{(}\\text{NO}_{3}\\text{)}_{3}}}\\times\\frac{{\\text{0.750 }\\cancel{{\\text{mol Pb}\\text{(}\\text{NO}_{3}\\text{)}_{3}}}}}{\\text{1000 }\\cancel{{\\text{mL Pb}\\text{(}\\text{NO}_{3}\\text{)}_{3}}}}\\times\\frac{\\text{2 }{\\cancel{\\text{mol NaI}}}}{\\text{1 }\\cancel{{\\text{mol Pb}\\text{(}\\text{NO}_{3}\\text{)}_{3}}}}\\times\\frac{\\text{1}}{\\text{0.0200}\\cancel{\\text{ L NaI}}}=\\text{4.5}\\text{ M NaI}[\/latex]<\/p>\n<p>3.<\/p>\n<p>(a). 2.15 g Al(OH)<sub>3<\/sub><\/p>\n<p style=\"text-align: center;\">[latex]\\text{55.0 }\\cancel{\\text{mL KOH}}\\times\\frac{{\\text{1.50 }\\cancel{{\\text{mol KOH}}}}}{\\text{1000 }\\cancel{\\text{mL KOH}}}\\times\\frac{\\text{2 }{\\cancel{\\text{mol Al}\\text{(OH)}_{3}}}}{\\text{6 }\\cancel{\\text{mol KOH}}}\\times\\frac{\\text{78.00 }\\text{g Al}\\text{(OH)}_{3}}{\\text{1 }\\cancel{\\text{mol Al}\\text{(OH)}_{3}}}=\\text{2.15 }\\text{g}\\text{ Al}\\text{(OH)}_{3}[\/latex]<\/p>\n<p>(b). 20.0 mL Al<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3<\/sub><\/p>\n<p style=\"text-align: center;\">[latex]\\text{10.0 }\\cancel{\\text{mL KOH}}\\times\\frac{{\\text{3.00 }\\cancel{{\\text{mol KOH}}}}}{\\text{1000 }\\cancel{\\text{mL KOH}}}\\times\\frac{\\text{1 }{\\cancel{\\text{mol Al}_{2}\\text{(SO}_{4}\\text{)}_{3}}}}{\\text{6 }\\cancel{\\text{mol KOH}}}\\times\\frac{\\text{1000 }\\text{mL Al}_{2}\\text{(SO}_{4}\\text{)}_{3}}{\\text{0.250 }\\cancel{\\text{mol Al}_{2}\\text{(SO}_{4}\\text{)}_{3}}}=\\text{20.0 }\\text{mL Al}_{2}\\text{(SO}_{4}\\text{)}_{3}[\/latex]<\/p>\n<p>(c). 1.50 M KOH<\/p>\n<p style=\"text-align: center;\">[latex]\\text{20.0 }\\cancel{\\text{mL Al}_{2}\\text{(SO}_{4}\\text{)}_{3}}\\times\\frac{{\\text{0.500 }\\cancel{\\text{mol Al}_{2}\\text{(SO}_{4}\\text{)}_{3}}}}{\\text{1000 }\\cancel{\\text{mL Al}_{2}\\text{(SO}_{4}\\text{)}_{3}}}\\times\\frac{\\text{6 }{\\cancel{\\text{mol KOH}}}}{\\text{1 }\\cancel{\\text{mol Al}_{2}\\text{(SO}_{4}\\text{)}_{3}}}\\times\\frac{\\text{1 }}{\\text{0.0400 }\\text{L Al}_{2}\\text{(SO}_{4}\\text{)}_{3}}=\\text{1.50 }\\text{M KOH}[\/latex]<\/p>\n<p>4. 0.184 L HCl solution<\/p>\n<p style=\"text-align: center;\">[latex]\\text{33.0 }\\cancel{\\text{g CuO}}\\times \\frac{\\text{1 }\\cancel{\\text{mol CuO}}}{\\text{79.55 }\\cancel{\\text{g CuO}}}\\times \\frac{\\text{2 }\\cancel{\\text{mol HCl}}}{\\text{1 }\\cancel{\\text{mol CuO}}}\\times \\frac{\\text{1 }\\cancel{\\text{L solution}}}{\\text{4.50 }\\cancel{\\text{mol HCl}}}=\\text{0.184 }\\text{L HCl}{\\text{ solution}}[\/latex]<\/p>\n<p>5. 2.80 L NO<sub>2<\/sub><\/p>\n<p style=\"text-align: center;\">[latex]\\text{0.150 }\\cancel{{\\text{L HNO}_{3}}}\\times\\frac{{\\text{0.500 }\\cancel{{\\text{mol HNO}_{3}}}}}{\\text{1 }\\cancel{\\text{L HNO}_{3}}}\\times\\frac{\\text{3 }{\\text{mol N}\\text{O}_{2}}}{\\text{2 }\\cancel{\\text{mol HCl}}}=\\text{0.113 mol}\\text{ NO}_{2}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{V}=\\frac{nRT}{P}=\\frac{\\left(\\text{0.113 }\\cancel{\\text{mol NO}_{2}}\\right){(0.0821\\large\\frac{\\text{L}\\cdot \\cancel{\\text{atm}}}{\\cancel{\\text{K}} \\cdot \\cancel{\\text{mol}}})}\\left(\\text{298 }\\cancel{\\text{K}}\\right)}{(\\text{750 }\\cancel{\\text{mmHg}}\\times\\large\\frac{\\text{1 }\\cancel{\\text{atm}}}{\\text{760 }\\cancel{\\text{mmHg}}})}=\\text{2.80 }\\text{L}\\text{ NO}_{2}[\/latex]<\/p>\n<p>6. 1.67 M HCl<\/p>\n<p style=\"text-align: center;\">[latex]{n}=\\frac{PV}{RT}=\\frac{(\\text{740 }\\cancel{\\text{torr}}\\times{}\\frac{\\text{1 }\\cancel{\\text{atm}}}{\\text{760 }\\cancel{\\text{torr}}})(\\text{0.720 }\\cancel{\\text{L}}\\text{ H}_{2})}{(0.0821\\large\\frac{\\cancel{\\text{L}}\\cdot \\cancel{\\text{atm}}}{\\cancel{\\text{K}} \\cdot \\text{mol}})(\\text{297 }\\cancel{\\text{K}})}=\\text{0.0288}\\text{ mol}\\text{ H}_{2}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\text{0.0288 }\\cancel{{\\text{mol H}_{2}}}\\times\\frac{\\text{2 }{\\text{mol HCl}}}{\\text{1 }\\cancel{\\text{mol H}_{2}}}\\times\\frac{1}{\\text{0.0344 }\\text{L solution}}=\\text{1.67 M}\\text{ HCl}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1496\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Solution Stoichiometry . <strong>Authored by<\/strong>:       Paul R. Young, Professor of Chemistry, University of Illinois at Chicago, Wiki: AskTheNerd; PRYufe6baskthenerd.com - pyoungufe6buic.edu; ChemistryOnline.com, Marisa Alviar-Agnew (Sacramento City College), Henry Agnew (UC Davis). <strong>Provided by<\/strong>: Libre Text. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Introductory_Chemistry\/Map%3A_Introductory_Chemistry_(Tro)\/13%3A_Solutions\/13.08%3A_Solution_Stoichiometry\">https:\/\/chem.libretexts.org\/Textbook_Maps\/Introductory_Chemistry\/Map%3A_Introductory_Chemistry_(Tro)\/13%3A_Solutions\/13.08%3A_Solution_Stoichiometry<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":6181,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Solution Stoichiometry \",\"author\":\"      Paul R. Young, Professor of Chemistry, University of Illinois at Chicago, Wiki: AskTheNerd; PRYufe6baskthenerd.com - pyoungufe6buic.edu; ChemistryOnline.com, Marisa Alviar-Agnew (Sacramento City College), Henry Agnew (UC Davis)\",\"organization\":\"Libre Text\",\"url\":\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Introductory_Chemistry\/Map%3A_Introductory_Chemistry_(Tro)\/13%3A_Solutions\/13.08%3A_Solution_Stoichiometry\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1496","chapter","type-chapter","status-publish","hentry"],"part":541,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1496","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/users\/6181"}],"version-history":[{"count":69,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1496\/revisions"}],"predecessor-version":[{"id":1975,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1496\/revisions\/1975"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/parts\/541"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1496\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/media?parent=1496"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=1496"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/contributor?post=1496"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/license?post=1496"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}