Produce dichlorine oxide by mixing sulphur dioxide and chlorine


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> Dichlorine oxide is used as bactericide to purify water. It is produced by the chlorination of sulfur dioxide gas.
>  $$\ce{SO2(g) + 2Cl2(g) -> SOCl2(l) + Cl2O(g)}$$ 
>  How many liters of dichlorine oxide ($\ce{Cl2O}$) can be produced by mixing 5.85L of $\ce{SO2}$ and 9.00L of $\ce{Cl2}$? 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.
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**My attempt:**


I just finished working out the problem and was hoping to get some feedback on my method of solving.


1. Convert both reactants from liters to moles using the ideal gas law and identify which is the limiting reactant.


n=$\frac{PV}{RT}$ = $\frac{\pu{5.85L}}{\pu{0.0821m/L}}$ = $71.2m \ce{SO2}$ (Constant T,P)


n=$\frac{PV}{RT}$ = $\frac{\pu{9.00L}}{\pu{0.0821m/L}}$ = 110m $\ce{Cl2}$ (Constant T,P)


$\ce{$71.2$m \ce{SO2 }X}\frac{1m\ce{Cl2O}}{1m\ce{SO2}} = 71.2m \ce{Cl2O}$ Excess Reactant


$110m \ce{Cl2}$ x $\frac{1m \ce{Cl2O}}{2m \ce{Cl2}}$ = $55m \ce{Cl2O}$ Limiting Reactant


2. Find amount of liters of $\ce{Cl2O}$ that can be made in the reaction by converting limiting reactant from moles to liters using the ideal gas law.


V=$\frac{nRT}{P}$ = $55m \ce{Cl2O}$ X 0.0821m/L = 4.5L $\ce{Cl2O}$ (Constant T,P)


4.5L of $\ce{Cl2O}$ can be produced by mixing 5.85L of $\ce{SO2}$ with 9.00L of $\ce{Cl2}$. 


3. Find liters of excess reactant left over after reaction by finding number of moles of excess reactant that can react with limiting reactant and then subtracting number from original amount of excess.


$110m \ce{Cl2}$ X $\frac{1m \ce{SO2}}{2m \ce{Cl2}}$ = $55m \ce{SO2}$


$71.2m \ce{SO2}$ - $55m \ce{SO2}$ = $16.2m \ce{SO2}$ of excess left over after reaction. 


