Question

Hydrogen,H2,and nitrogen,N2(g), combined to form ammonia,NH3(g):3H2+N2–>2NH3(g) what amount, in moles, of nitrogen will react 18 mols of hydrogen

Answer 1

Answer:

In this reaction, $6\; \rm mol$ of $\rm N_2$  will react with $18\; \rm mol$ of $\rm H_2$.

Explanation:

In the balanced equation for this reaction, the ratio between the coefficient of $\rm H_2$ and $\rm N_2$ is $3 : 1$. That is: $n(\mathrm{H_2}) : n(\mathrm{N_2}) = 3:1$. That's the same as saying that for every one mole of $\rm N_2$ consumed, three moles of $\rm H_2$ will be consumed.

Rewrite this ratio as a fraction:

$\displaystyle \frac{n(\mathrm{H_2})}{n(\mathrm{N_2})} = \frac{3}{1}$.

Take the reciprocal of both sides to obtain:

$\displaystyle \frac{n(\mathrm{N_2})}{n(\mathrm{H_2})} = \frac{1}{3}$.

It is given that $18\; \rm mol$ of $\rm H_2$ was consumed; in other words, $n(\mathrm{H_2}) = 18\; \rm mol$. The question is asking for $n(\mathrm{N_2})$, the number of moles of $\rm N_2$ required. Apply this ratio:

$\begin{aligned}n(\mathrm{N_2}) &= n(\mathrm{H_2}) \cdot \frac{n(\mathrm{N_2})}{n(\mathrm{H_2})}\\ &= 18\; \rm mol \times \frac{1}{3} = 6\; \rm mol\end{aligned}$.

Hence the conclusion: in this reaction, it will take (at least) $6\; \rm mol$ of $\rm N_2$ to react with $18\; \rm mol$ of $\rm H_2$.