# Question 827fe

Nov 9, 2017

The theoretical yield of this reaction is $5.20 g$ of $H I$. By extension, the percent yielded here is 60.8%; acceptable, but it could be improved.
${H}_{2} \left(g\right) + {I}_{2} \left(s\right) \to 2 H I \left(g\right)$
$5.16 g \cdot \frac{{I}_{2}}{253.8 g} \cdot \frac{2 H I}{{I}_{2}} \cdot \frac{127.9 g}{H I} \approx 5.20 g$
(3.16g)/(5.20g) approx 60.8%#