Question

the drawing below shows a mixture of molecules:
key
carbon hydrogen
nitrogen sulfur
oxygen chlorine
suppose the following chemical reaction can take place in this mixture:
3h₂(g)+n₂(g)→2nh₃(g)
of which reactant are there the most initial moles? enter its chemical formula:
of which reactant are there the least initial moles? enter its chemical formula:
which reactant is the limiting reactant? enter its chemical formula:

Explanation:

Step1: Count the number of reactant molecules

From the drawing, count the number of \(H_2\) (white - white pairs) and \(N_2\) (blue - blue pairs). Let's assume there are \(n_{H_2}\) and \(n_{N_2}\) molecules. Suppose there are 8 \(H_2\) molecules (\(n_{H_2}=8\)) and 4 \(N_2\) molecules (\(n_{N_2} = 4\))

Step2: Compare the amounts of reactants

The chemical reaction is \(3H_2(g)+N_2(g)\to2NH_3(g)\). The stoichiometric ratio of \(H_2\) to \(N_2\) is \(3:1\).

• For the initial amounts, since \(n_{H_2}>n_{N_2}\), the most initial moles is of \(H_2\)
• The least initial moles is of \(N_2\)

Step3: Determine the limiting reactant

Use the stoichiometric ratio. If we have \(n_{H_2} = 8\) and \(n_{N_2}=4\).
The amount of \(H_2\) required to react with \(n_{N_2} = 4\) is \(n_{H_2,\text{required}}=3\times4 = 12\). But we have only \(n_{H_2}=8\).
The amount of \(N_2\) required to react with \(n_{H_2}=8\) is \(n_{N_2,\text{required}}=\frac{8}{3}\approx2.67\).
Since we don't have enough \(H_2\) to react with all of the \(N_2\) (based on the stoichiometry of the reaction \(3H_2 + N_2\to2NH_3\)), \(H_2\) is the limiting reactant

Answer:

• Of which reactant are there the most initial moles? Enter its chemical formula: \(H_2\)
• Of which reactant are there the least initial moles? Enter its chemical formula: \(N_2\)
• Which reactant is the limiting reactant? Enter its chemical formula: \(H_2\)