# Question 93cb7

Apr 20, 2016

${H}_{2}$ is the limiting reactant.
10 moles of $H B r$ are produced.
160 g of $B {r}_{2}$ are remaining after reaction.

#### Explanation:

Let us write again the balanced equation:

${H}_{2} \left(g\right) + B {r}_{2} \left(g\right) \to 2 H B r \left(g\right)$

In order to find the amount of $H B r \left(g\right)$ that will form we will need to know which one is the limiting reactant.

$\textcolor{g r e e n}{\text{Using 5mol } {H}_{2}}$:

?molHBr=5cancel(molH_2)xx(2molHBr)/(1cancel(molH_2))=color(green)(10molHBr)

color(red)("Using 7mol "Br_2:

?molHBr=7cancel(molBr_2)xx(2molHBr)/(1cancel(molBr_2))=color(red)(14molHBr)#

In general, the reactant that produces the least amount of $H B r$ is the limiting reactant. Therefore, ${H}_{2}$ is the $\textcolor{g r e e n}{\text{limiting reactant}}$ and $B {r}_{2}$ is the $\textcolor{red}{\text{excess reactant}}$.

Since $\textcolor{g r e e n}{1 m o l {H}_{2}}$ reacts with $\textcolor{red}{1 m o l B {r}_{2}}$, therefore, $\textcolor{g r e e n}{5 m o l {H}_{2}}$ will react only with $\textcolor{red}{5 m o l B {r}_{2}}$ and the excess amount of $B {r}_{2}$ is $\textcolor{red}{2 m o l B {r}_{2}}$.

In order to find the mass in grams for the excess $B {r}_{2}$ remaining after reaction:

$n = \frac{m}{M M} \implies m = n \times M M$

$m = 2 \cancel{m o l B {r}_{2}} \times \frac{80 g B {r}_{2}}{1 \cancel{m o l B {r}_{2}}} = 160 g B {r}_{2}$