Question

nitrogen (n₂) gas and hydrogen (h₂) gas react to form ammonia (nh₃) gas. suppose you have 9.0 mol of n₂ and 1.0 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 \(NH_3\) that can be formed from each reactant

• From \(N_2\):

Using the mole ratio from the balanced equation (\(\frac{2\ mol\ NH_3}{1\ mol\ N_2}\)), if we have \(n(N_2) = 9.0\ mol\), then \(n(NH_3)\) from \(N_2\) is \(n(NH_3)=9.0\ mol\times\frac{2\ mol\ NH_3}{1\ mol\ N_2}=18.0\ mol\)

• From \(H_2\):

Using the mole ratio (\(\frac{2\ mol\ NH_3}{3\ mol\ H_2}\)), if \(n(H_2)=1.0\ mol\), then \(n(NH_3)\) from \(H_2\) is \(n(NH_3)=1.0\ mol\times\frac{2\ mol\ NH_3}{3\ mol\ H_2}=\frac{2}{3}\ mol\approx0.67\ mol\)

Step3: Identify the limiting reactant

The reactant that produces the least amount of product is the limiting reactant. Since \(H_2\) produces less \(NH_3\) (\(0.67\ mol\) compared to \(18.0\ mol\) from \(N_2\))

Answer:

\(H_{2}\)