### Learning Objectives

By the end of this section, you will be able to:

- Convert between units of mass, mol, and between substances using stoichiometry.
- Predict and balance a reaction of calcium.
- Utilize proper laboratory techniques.

**Introduction**

Stoichiometry is the study of the mathematical relationships between chemicals. These relationships are set up as dimensional analysis problems where the units cancel until you are left with the unit of the answer. For example, in lecture you have discussed the conversion of grams to mol using molar mass. If you were to weigh out 5.0 grams of Ca, it is possible to convert it to mol using the molar mass found on the periodic table.

[latex]\frac{5\text{g Ca}}{\text{}}\,\,\,\,\,\,\,\,\frac{1\text{ mol Ca}}{40.08\text{ g Ca}}=0.12\text{ mol Ca}[/latex]

Notice the answer is given to the correct number of significant figures. You have also investigated the conversion between mol and either atoms or molecules using Avogadro’s number in lecture. It is possible to convert between mol of one substance and mol of another using the coefficients in the balanced equation. For example the balanced equation between sulfuric acid and NaOH is given below.

[latex]\text{H}_{2}\text{SO}_{4}+2\text{NaOH}{\rightarrow}\text{Na}_{2}\text{SO}_{4}+\text{H}_{2}\text{O}[/latex]

This balanced equation indicates that 1 mol of sulfuric acid reacts with 2 mol of sodium hydroxide to produce 1 mol of sodium sulfate and 2 mol of water. It is possible to derive from this that every time 1 mol of sulfuric acid reacts, 2 mol of sodium hydroxide are needed. Using this we can determine how many mol of sodium hydroxide are needed to react with 24.0 g of sulfuric acid. We will have to convert first to mol sulfuric acid using the molar mass sulfuric acid, then to mol sodium hydroxide using the coefficients in the balanced equation.

[latex]\text{g H}_{2}\text{SO}_{4}{\rightarrow}\text{mol H}_{2}\text{SO}_{4}{\rightarrow}\text{mol }\text{NaOH}[/latex]

[latex]\text{}\frac{24.0\text{g H}_2\text{SO}_4}{\text{}}\,\,\,\,\,\,\,\,\frac{1\text{ mol H}_2\text{SO}_4}{98.09\text{ H}_2\text{SO}_4}\,\,\,\,\,\,\,\,\frac{2\text{mol NaOH}}{1\text{ mol H}_2\text{SO}_4}=489\text{ mol NaOH}[/latex]

Figure 1 summarizes some of the stoichiometric relationships that are used this semester. These relationships are investigated in detail in this lab as we look at reactions of calcium. Students will look at how calcium reacts with water and hydrochloric acid. Students will also evaluate the reaction between hydrochloric acid and sodium hydroxide. The stoichiometric relationship between hydrochloric acid and calcium will be determined both graphically and experimentally.