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 21.0 mol of H2 and excess N2?

Answer 1

14 moles of NH3 can be produced from 21.0 mol of H2 and excess N2.

Explanation:

Balanced equation for the production of ammonia is given as:

3H2(g)+N2(g)→2NH3(g)

Given that:

number of moles of H2 = 21 moles

moles of NH3 produced from 21 moles of H2 =?

In the above equation the limiting reagent is hydrogen gas so the production of ammonia directly depends on it.

From the chemical equation we can see that:

3 moles of H2 react to form 2 moles of NH3

so, 21 moles of  H2 will react to form x moles of NH3

$\frac{2}{3}$ = $\frac{x}{21}$

= $\frac{21 X 2}{3}$

= 14 MOLES

from the calculation by applying stoichiometry it is found that 14 moles of ammonia is produced by using 21 moles of hydrogen.