Question

Consider the following balanced reaction between hydrogen and nitrogen to form ammonia:3H2(g) + N2(g)→2NH3(g)How many moles of NH3 can be produced from 24.0 mol of H2 and excess N2?Express the number of moles to three significant figures.

Answer 1

Answer:

16.0 moles

Explanation:

The reaction is:

$3H_2_{(g)}+N_2_{(g)}\rightarrow 2NH_3_{(g)}$

It is given that $N_2$ is present in large excess. It means that $H_2$ is the limiting reagent and the formation of the product is governed by $H_2$.

From the reaction stoichiometry,

3 moles of $H_2$ on reaction forms 2 moles of $NH_3$

Also,

1 mole of $H_2$ on reaction forms 2/3 moles of $NH_3$

Given moles = 24.0 moles

24 moles of $H_2$ on reaction forms (2/3)×24.0 moles of $NH_3$

Moles of $NH_3$ = 16.0 moles