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

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

Step-by-step 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})} = (3)/(1)`.

Take the reciprocal of both sides to obtain:

`\displaystyle \frac{n(\mathrm{N_2})}{n(\mathrm{H_2})} = (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 * (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`.