# What is my actual yield of uranium hexabromide if start with 100 grams of uranium and get a percent yield of 83%?

## $U + 3 B {r}_{2} \to U B {r}_{6}$

Jan 1, 2018

Assuming that uranium is the limiting reactant, and the equation you stated is correct,

$U + 3 B {r}_{2} \to U B {r}_{6}$

The theoretical yield is,

$100 g \cdot \frac{U}{238 g} \cdot \frac{U B {r}_{6}}{U} \approx 0.420 \text{mol}$,

and if the percent yield is 83%,

$0.83 = \frac{A}{0.420}$
$\therefore A \approx 0.350 \text{mol}$

is the actual yield of uranium hexabromide given this data.