Question

How many moles of NH3 could you produce with 0.24 moles of H2 and excess nitrogen gas?

Answer 1

`0.16\; \rm mol`.

Step-by-step explanation:

`\rm H_2` and
`\rm N_2` react to produce
`\rm NH_3` through the following reaction:

`\rm 2\; N_2 + 3\; H_2 \to 2\; NH_3`.

The ratio between the coefficients and
`\rm NH_3` and
`\rm H_2` is:

`\displaystyle \frac{n({\rm NH_3})}{n({\rm H_2})} = (2)/(3)`.

This coefficient ratio would be the ratio between the quantity of
`\rm NH_3` produced and the quantity of
`\rm H_2` consumed in this reaction only if
`\rm H_2\!` is a limiting reactant.

The other reactant in this reaction,
`\rm N_2`, is in excess. Hence,
`\rm H_2` would be the limiting reactant. Hence, the coefficient ratio could be used to find the quantity of
`\rm NH_3` produced in this reaction.

`\begin{aligned}n({\rm NH_3}) &= \frac{n({\rm NH_3})}{n({\rm H_2})} \cdot n({\rm H_2}) \\ &= (2)/(3) * 0.24\; \rm mol \\ &= 0.16\; \rm mol \end{aligned}`.