Question

if 10.0 moles of o₂ are reacted with excess no in the reaction below, and 3.7 mol of no₂ are collected, then what is the percent yield for the reaction? 2no(g) + o₂(g) → 2no₂(g)

Explanation:

Step1: Determine theoretical yield

From the balanced equation $2NO(g)+O_2(g)\to 2NO_2(g)$, the mole - ratio of $O_2$ to $NO_2$ is 1:2. Given 10.0 moles of $O_2$, the theoretical amount of $NO_2$ produced is $n_{theo}=10.0\ mol\ O_2\times\frac{2\ mol\ NO_2}{1\ mol\ O_2}=20.0\ mol\ NO_2$.

Step2: Calculate percent yield

The percent - yield formula is $\text{Percent Yield}=\frac{n_{actual}}{n_{theo}}\times 100\%$. Given $n_{actual} = 3.7\ mol$ and $n_{theo}=20.0\ mol$, then $\text{Percent Yield}=\frac{3.7\ mol}{20.0\ mol}\times 100\% = 18.5\%$.

Answer:

18.5%