Question

nitroglycerin (c₃h₅n₃o₉) is a powerful explosive, used in dynamite. it decomposes according to the reaction:
4 c₃h₅n₃o₉ → 6 n₂ + 12 co₂ + 10 h₂o + o₂
this reaction generates a lot of heat, and this, along with the sudden fast formation and rapid expansion of the gases produced, is what causes the explosion.
if one begins with 650. g of nitroglycerin and (somehow) recovers 20.7 g of o₂, what is the percent yield of this reaction? report your answer to 3 significant figures.

Explanation:

Step1: Calculate the molar mass of nitroglycerin ($C_3H_5N_3O_9$)

The molar mass of $C$ is $12.01\ g/mol$, $H$ is $1.008\ g/mol$, $N$ is $14.01\ g/mol$, and $O$ is $16.00\ g/mol$.
$M = 3\times12.01+5\times1.008 + 3\times14.01+9\times16.00$
$M=36.03 + 5.04+42.03+144.00$
$M = 227.1\ g/mol$

Step2: Calculate the moles of nitroglycerin

Use the formula $n=\frac{m}{M}$, where $m = 650. g$ and $M = 227.1\ g/mol$
$n=\frac{650.}{227.1}\approx2.862\ mol$

Step3: Determine the moles of $O_2$ produced based on stoichiometry

From the balanced equation $4C_3H_5N_3O_9
ightarrow6N_2 + 12CO_2+10H_2O+O_2$, the mole ratio of $C_3H_5N_3O_9$ to $O_2$ is $4:1$
So moles of $O_2$ produced (theoretical) $n_{O_2}=\frac{2.862}{4}=0.7155\ mol$

Step4: Calculate the theoretical mass of $O_2$

The molar mass of $O_2$ is $M_{O_2}=32.00\ g/mol$
Theoretical mass $m_{theo}=n_{O_2}\times M_{O_2}=0.7155\times32.00 = 22.9\ g$

Step5: Calculate the percent yield

Percent yield formula is $\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%$
Actual yield $m_{actual}=20.7\ g$, theoretical yield $m_{theo}=22.9\ g$
$\text{Percent Yield}=\frac{20.7}{22.9}\times100\%\approx90.4\%$

Answer:

$90.4\%$