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

User Rob Bonner
by
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1 Answer

3 votes

Answer:


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}.

User Justin G
by
6.2k points