Question

In a closed container, 4.0 moles of N were allowed to react with 9.0 moles of hydrogen gas. The analysis of the mixture of gases in the container after some time revealed the presence of 3.0 moles of ammonia. Using the information above answer the question below.What is the actual yield of the reaction in moles (the amount of the product that was actually produced)?

Answer 1

The reaction is:

`\text{ N}_(2(g))\text{ + 3H}_(2(g))\text{ }\rightarrow\text{ 2NH}_(3(g))`
`4\text{ mol N}_2\text{ }*\frac{2\text{ mol NH}_3}{1\text{ mol N}_2}=8\text{ mol NH}_3`
`9\text{ mol H}_2*\frac{2\text{ mol NH}_3}{3\text{ mol H}_2}=\text{ 6 mol NH}_3`

The limiting reactant is hydrogen and theoretically 6 moles of ammonia would be produced

`\begin{gathered} \text{ }\%\text{ yield =}\frac{\text{ actual yield}}{\text{ theoretical yield }}*100\% \\ \%\text{ yield =}\frac{3\text{ mol}}{6\text{ mol }}*100\%\text{ = 50}\% \end{gathered}`

So actual yield = 3 mol NH3

Theoretical yield = 6 mol NH3