Question

nitrogen (n₂) gas and hydrogen (h₂) gas react to form ammonia (nh₃) gas. suppose you have 11.0 mol of n₂ and 7.0 mol of h₂ in a reactor. what would be the limiting reactant? enter its chemical formula below.

Explanation:

Step1: Write the balanced chemical equation

The reaction between \(N_2\) and \(H_2\) to form \(NH_3\) is \(N_{2}(g)+3H_{2}(g)
ightarrow 2NH_{3}(g)\)

Step2: Calculate the moles of \(N_2\) required if \(H_2\) is completely consumed

From the balanced equation, the mole ratio of \(N_2\) to \(H_2\) is \(1:3\). If \(n(H_2) = 7.0\space mol\), then the moles of \(N_2\) required \(n_{req}(N_2)=\frac{1}{3}\times n(H_2)\)
\(n_{req}(N_2)=\frac{1}{3}\times7.0\space mol\approx2.33\space mol\)
Since we have \(n(N_2) = 11.0\space mol\) (which is more than \(2.33\space mol\)), \(H_2\) is not in excess.

Step3: Calculate the moles of \(H_2\) required if \(N_2\) is completely consumed

From the mole ratio (\(N_2:H_2 = 1:3\)), if \(n(N_2)=11.0\space mol\), then the moles of \(H_2\) required \(n_{req}(H_2)=3\times n(N_2)\)
\(n_{req}(H_2)=3\times11.0\space mol = 33.0\space mol\)
But we have only \(n(H_2)=7.0\space mol\). So, \(H_2\) will be consumed first.

Answer:

\(H_{2}\)