Answer:
[tex]0.16\; \rm mol[/tex].
Explanation:
[tex]\rm H_2[/tex] and [tex]\rm N_2[/tex] react to produce [tex]\rm NH_3[/tex] through the following reaction:
[tex]\rm 2\; N_2 + 3\; H_2 \to 2\; NH_3[/tex].
The ratio between the coefficients and [tex]\rm NH_3[/tex] and [tex]\rm H_2[/tex] is:
[tex]\displaystyle \frac{n({\rm NH_3})}{n({\rm H_2})} = \frac{2}{3}[/tex].
This coefficient ratio would be the ratio between the quantity of [tex]\rm NH_3[/tex] produced and the quantity of [tex]\rm H_2[/tex] consumed in this reaction only if [tex]\rm H_2\![/tex] is a limiting reactant.
The other reactant in this reaction, [tex]\rm N_2[/tex], is in excess. Hence, [tex]\rm H_2[/tex] would be the limiting reactant. Hence, the coefficient ratio could be used to find the quantity of [tex]\rm NH_3[/tex] produced in this reaction.
[tex]\begin{aligned}n({\rm NH_3}) &= \frac{n({\rm NH_3})}{n({\rm H_2})} \cdot n({\rm H_2}) \\ &= \frac{2}{3} \times 0.24\; \rm mol \\ &= 0.16\; \rm mol \end{aligned}[/tex].
