1. From the ideal gas law, PV = nRT, set [latex]\large n=\frac{\text{mass}}{\text{molar mass}}[/latex] and solve the molar mass. [latex]\text{molar mass}=\frac{\left(\text{0.281 g}\right)\left(0.08206\cancel{\text{L}}\cancel{\text{atm}}{\text{mol}}^{-\text{1}}\cancel{{\text{K}}^{-\text{1}}}\right)\left(399.15\cancel{\text{K}}\right)}{\left(\frac{777\cancel{\text{torr}}}{760\cancel{\text{torr}}\cancel{{\text{atm}}^{-\text{1}}}}\right)\left(0.125\cancel{\text{L}}\right)}=\text{72.0 g}{\text{mol}}^{-\text{1}}[/latex]

3. The answers are as follows:

1. Determine the moles of HgO that decompose; using the chemical equation, determine the moles of O₂ produced by decomposition of this amount of HgO; and determine the volume of O₂ from the moles of O₂, temperature, and pressure.
2. [latex]\large\begin{array}{l}\\ \\ 5.36\cancel{\text{g HgO}}\times \frac{\text{1 mol HgO}}{\left(200.59+15.9994\right)\cancel{\text{g HgO}}}=\text{0.0247 mol HgO}\\ 0.0247\cancel{\text{mol HgO}}\times \frac{\text{1 mol}{\text{O}}_{2}}{2\cancel{\text{mol HgO}}}=\text{0.01235 mol}{\text{O}}_{2}\end{array}[/latex]

PV = nRT
P = 0.975 atm
T = (23.0 + 273.15) K

[latex]\large V=\frac{nRT}{P}=\frac{0.01235\cancel{\text{mol}}\left(\text{0.08206 L}\cancel{\text{atm}}\cancel{{\text{mol}}^{-\text{1}}}\cancel{{\text{K}}^{-\text{1}}}\right)\left(296.15\cancel{\text{K}}\right)}{0.975\cancel{\text{atm}}}=0.308\text{ L}[/latex]

5. The answers are as follows:

1. Determine the molar mass of CCl₂F₂. From the balanced equation, calculate the moles of H₂ needed for the complete reaction. From the ideal gas law, convert moles of H₂ into volume.
2. Molar mass of CCl₂ F₂ = 12.011 + 2 × 18.9984 + 2 × 35.4527 = 120.913 g/mol
   [latex]\large\text{mol}{\text{H}}_{2}=1.000\times {10}^{6}\text{g}\times \frac{\text{1 mol}{\text{CCL}}_{2}{\text{F}}_{2}}{\text{120.913 g}}\times \frac{\text{4 mol}{\text{H}}_{2}}{\text{1 mol}{\text{CCl}}_{2}{\text{F}}_{2}}=3.308\times {10}^{4}\text{mol}[/latex]
   [latex]\large V=\frac{nRT}{P}=\frac{\left(3.308\times {10}^{4}\cancel{\text{mol}}\right)\left(\text{0.08206 L}\cancel{\text{atm}}\cancel{{\text{mol}}^{-\text{1}}}{\cancel{\text{K}}}^{-\text{1}}\right)\left(308.65\cancel{\text{K}}\right)}{225\cancel{\text{atm}}}=3.72\times {10}^{3}\text{L}[/latex]

7. The answers are as follows:

1. Balance the equation. Determine the grams of CO₂ produced and the number of moles. From the ideal gas law, determine the volume of gas.
2. [latex]\large{\text{CaCO}}_{3}\left(s\right)\rightarrow\text{CaO}\left(s\right)+{\text{CO}}_{2}\left(g\right)[/latex]
   [latex]\large\text{mass}{\text{CO}}_{2}=1.00\times {10}^{6}\text{g}\times \frac{\text{1 mol}{\text{CaCO}}_{2}}{\text{100.087 g}}\times \frac{\text{44.01 g}{\text{CO}}_{2}}{\text{1 mol}{\text{CO}}_{2}}\times \frac{\text{1 mol}{\text{CO}}_{2}}{\text{1 mol}{\text{CaCO}}_{2}}=4.397\times {10}^{5}\text{g}[/latex]
   [latex]\large\text{mol}{\text{CO}}_{2}=\frac{4.397\times {10}^{5}\text{g}}{\text{44.01 g}{\text{mol}}^{-\text{1}}}=\text{9991 mol}[/latex]
   [latex]V=\frac{nRT}{P}=\frac{\left(\text{9991 mol}\right)\left(\text{0.08206 L atm}{\text{mol}}^{-\text{1}}{\text{K}}^{-\text{1}}\right)\left(\text{875 K}\right)}{\text{0.966 atm}}=7.43\times {10}^{5}\text{L}[/latex]

9. [latex]\large 2{\text{C}}_{2}{\text{H}}_{6}\left(\text{g}\right)+7{\text{O}}_{2}\left(\text{g}\right)\rightarrow 4{\text{CO}}_{2}\left(\text{g}\right)+6{\text{H}}_{2}\text{O}\left(\text{g}\right)[/latex]

From the balanced equation, we see that 2 mol of C₂H₆ requires 7 mol of O₂ to burn completely. Gay-Lussac’s law states that gases react in simple proportions by volume. As the number of liters is proportional to the number of moles,

[latex]\large\frac{\text{12.00 L}}{\text{2 mol}{\text{C}}_{2}{\text{H}}_{6}}=\frac{V\left({\text{O}}_{2}\right)}{\text{7 mol}{\text{O}}_{2}}[/latex]
[latex]V\left({\text{O}}_{2}\right)=\frac{\text{12.00 L}\times 7}{2}=\text{42.00 L}[/latex]