Dichlorine oxide is used as bactericide to purify water. It is produced by the chlorination of sulfur dioxide gas. $$ \mathrm{SO}_{2}(g)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SOCl}_{2}(l)+\mathrm{Cl}_{2} \mathrm{O}(g) $$ How many liters of \(\mathrm{Cl}_{2} \mathrm{O}\) can be produced by mixing \(5.85 \mathrm{~L}\) of \(\mathrm{SO}_{2}\) and \(9.00 \mathrm{~L}\) of \(\mathrm{Cl}_{2}\) ? How many liters of the reactant in excess are present after reaction is complete? Assume \(100 \%\) yield and that all the gases are measured at the same temperature and pressure.

At constant temperature and pressure, the number of moles of a gas is directly proportional to its volume (Avogadro's Law). Therefore, we can use the ratio of the volume of each reactant to calculate the number of moles: mol SO2 = (5.85 L SO2) * (1 mol SO2 / 1 L SO2) = 5.85 mol SO2 mol Cl2 = (9.00 L Cl2) * (1 mol Cl2 / 1 L Cl2) = 9.00 mol Cl2

Compare the ratio of moles of reactants to the ratio of moles based on the balanced chemical equation: \( SO_2(g) + 2 Cl_2(g) \rightarrow SOCl_2(l) + Cl_2O(g)\) mol SO2 / mol Cl2 = 5.85 / 9.00 = 0.65 Since there are two moles of Cl2 needed for each mole of SO2, we need to see whether this ratio is less than or greater than 2. 0.65 < 2, so the ratio of moles indicates that SO2 will run out first. Thus, SO2 is the limiting reactant.

To calculate the volume of Cl2O produced, we first need to find the moles of Cl2O formed, using the stoichiometry of the balanced chemical equation: mol Cl2O = (mol SO2) * (1 mol Cl2O / 1 mol SO2) = 5.85 mol Cl2O Now, using Avogadro's Law, we convert moles of Cl2O to volume: Volume Cl2O = (5.85 mol Cl2O) * (1 L Cl2O / 1 mol Cl2O) = 5.85 L Cl2O

Since we know that SO2 is the limiting reactant, we need to find out how much Cl2 was used and how much is left over. To do this, we'll use the stoichiometry of the balanced chemical equation: mol Cl2 used = (mol SO2) * (2 mol Cl2 / 1 mol SO2) = 5.85 * 2 = 11.70 mol Cl2 Now, we subtract the moles of Cl2 used from the initial moles of Cl2: mol Cl2 left = 9.00 mol Cl2 - 11.70 mol Cl2 = -2.70 mol Cl2 Since we got a negative number, we made an error in our calculations. Re-checking the stoichiometry, we find that the ratio between the moles of SO2 and Cl2 should actually be 1:2, not 1:1. So we need to recalculate the moles of Cl2 used: mol Cl2 used = (mol SO2) * (2 mol Cl2 / 1 mol SO2) = 5.85 * 1 = 5.85 mol Cl2 Now, subtract the moles of Cl2 used from the initial moles of Cl2: mol Cl2 left = 9.00 mol Cl2 - 5.85 mol Cl2 = 3.15 mol Cl2 Finally, we convert the moles of excess Cl2 to volume: Volume Cl2 left = (3.15 mol Cl2) * (1 L Cl2 / 1 mol Cl2) = 3.15 L Cl2 So, the volume of Cl2O produced is 5.85 L, and the volume of excess Cl2 left over after the reaction is 3.15 L.