# What is the percent yield if 50.0 g of Cr is formed from the reaction of 150.0 g of Cr_2O_3 (MW = 152.0 amu) with excess Al?

Aug 25, 2016

Approx 50% yield

#### Explanation:

$C {r}_{2} {O}_{3} + 2 A l + \Delta \rightarrow 2 C r + A {l}_{2} {O}_{3}$

$\text{Moles of } C {r}_{2} {O}_{3}$ $=$ $\frac{150 \cdot g}{151.99 \cdot g \cdot m o {l}^{-} 1}$ $\cong 1 \cdot m o l$

$\text{Moles of } C r$ $=$ $\frac{50 \cdot g}{52.00 \cdot g \cdot m o {l}^{-} 1}$ $\cong 1 \cdot m o l$

This is a nasty question as it is trying to get you to ignore the given stoichiometry of the reaction, and report 100% yield. However, 100% yield would give $104 \cdot g$ chromium metal, because the decomposition yields 2 mol of chromium per mol of $C {r}_{2} {O}_{3}$.