Answer
22.4 $CS_{2}$
Work Step by Step
The strategy to find the mass of $CS_{2}$ that react with 62.7 g $Cl_{2}$ is:
g $Cl_{2}$ ${\rightarrow}$ mol $Cl_{2}$ ${\rightarrow}$ mol $CS_{2}$${\rightarrow}$ g $CS_{2}$
The conversion factors are:
$\frac{1 mol (Cl_{2})}{Molar mass (Cl_{2})}$ to convert to moles $Cl_{2}$, where Molar mass $Cl_{2}$:
Molar mass $Cl_{2}$ = 2 x 35.5 = 71.0 g
$\frac{1 mol (CS_{2})}{3 mol (Cl_{2})}$, to convert the moles of $Cl_{2}$ to mol of $CS_{2}$.
$\frac{Molar mass(CS_{2})}{1 mol(CS_{2})}$ to convert to g of $CS_{2}$, where the molar mass of $CS_{2}$ is:
Molar mass $CS_{2}$ = 2 x 12.0 + 2 x 32.0 = 76.0 g
Now we use the conversion factors to find the mas of $CS_{2}$:
62.7 g$Cl_{2}$ x $\frac{1 mol (Cl_{2})}{Molar mass (Cl_{2})}$ x $\frac{1 mol (CS_{2})}{3 mol (Cl_{2})}$x $\frac{Molar mass(CS_{2})}{1 mol(CS_{2})}$ =
62.7 g $Cl_{2}$ x $\frac{1 mol (Cl_{2})}{71 g (Cl_{2})}$ x $\frac{1 mol (CS_{2})}{3 mol (Cl_{2})}$x $\frac{76 g(CS_{2})}{1 mol(CS_{2})}$ = 22.37 g $CS_{2}$ or 22.4 $CS_{2}$
