Answer
5.7 x $10^{5}$ g $C_{3}H_{3}N$
Work Step by Step
Balanced chemical reaction:
4 $C_{3}H_{6}$ (g) + 6 NO (g) ${\longrightarrow}$ 4 $C_{3}H_{3}N$ (g) + 6 $H_{2}O$ (g) + $N_{2}$ (g)
The strategy to find the mass of $C_{3}H_{3}N$that can be obtained from 452 kg $C_{3}H_{6}$ is:
g $C_{3}H_{6}$ ${\rightarrow}$ mol $C_{3}H_{6}$ ${\rightarrow}$ mol $C_{3}H_{3}N$${\rightarrow}$ g $C_{3}H_{3}N$
The conversion factors are:
$\frac{1 mol (C_{3}H_{6})}{Molar mass (C_{3}H_{6})}$ to convert to moles $C_{3}H_{6}$, where Molar mass $C_{3}H_{6}$:
Molar mass $C_{3}H_{6}$ = 3 x 12.0 + 6 x 1.0 = 42 g
$\frac{4 mol (C_{3}H_{3}N)}{4 mol (C_{3}H_{6})}$, to convert the moles of $C_{3}H_{6}$ to mol of $C_{3}H_{3}N$.
$\frac{Molar mass(C_{3}H_{3}N)}{1 mol(C_{3}H_{3}N)}$ to convert to g of $C_{3}H_{3}N$, where the molar mass of $C_{3}H_{3}N$ is:
Molar mass $C_{3}H_{3}N$ = 3 x 12.0 + 3 x 1.0 + 1 x 14.0 = 53 g
Now we use of the conversion factors to find the mas of $C_{3}H_{3}N$:
452 x $10^{3}$ g $C_{3}H_{6}$ x $\frac{1 mol (C_{3}H_{6})}{Molar mass (C_{3}H_{6})}$ x $\frac{4 mol (C_{3}H_{3}N)}{4 mol (C_{3}H_{6})}$ x $\frac{Molar mass(C_{3}H_{3}N)}{1 mol(C_{3}H_{3}N)}$ =
452 x $10^{3}$ g $C_{3}H_{6}$ x $\frac{1 mol (C_{3}H_{6})}{42 g (C_{3}H_{6})}$ x $\frac{4 mol (C_{3}H_{3}N)}{4 mol (C_{3}H_{6})}$ x $\frac{53g (C_{3}H_{3}N)}{1 mol(C_{3}H_{3}N)}$ = 570 x $10^{3}$ g $C_{3}H_{3}N$ or
5.7 x $10^{5}$ g $C_{3}H_{3}N$
