Question

aluminum and oxygen react according to the following equation:
4 al(s) + 3 o₂(g) → 2 al₂o₃(s)
in a certain experiment, 0.170 mol al was reacted with excess oxygen to produce 0.067 mol of product. what was the percent yield of this reaction?
report your answer with two significant figures.

Explanation:

Step1: Determine the theoretical yield

From the balanced equation \(4Al(s)+3O_{2}(g)\to2Al_{2}O_{3}(s)\), the mole ratio of \(Al\) to \(Al_{2}O_{3}\) is \(4:2 = 2:1\).
If \(n(Al)=0.170\space mol\), then the theoretical amount of \(Al_{2}O_{3}\) produced, \(n_{theo}(Al_{2}O_{3})=\frac{0.170\space mol}{2}=0.085\space mol\)

Step2: Calculate the percent yield

The formula for percent yield is \(\text{Percent Yield}=\frac{n_{actual}}{n_{theo}}\times100\%\)
Given \(n_{actual}(Al_{2}O_{3}) = 0.067\space mol\) and \(n_{theo}(Al_{2}O_{3})=0.085\space mol\)
\(\text{Percent Yield}=\frac{0.067\space mol}{0.085\space mol}\times 100\% \)
\(\text{Percent Yield}=\frac{67}{85}\times 100\% \approx 79\%\)

Answer:

\(79\%\)