Question

how many moles of nh₃ can be produced by the reaction of 2.00 g of n₂ with 3.00 g h₂? reaction: n₂(g) + 3 h₂(g) → 2 nh₃(g) 0.235 0.0567 0.143 1.00 none of the above

Explanation:

Step1: Calculate moles of \(N_2\)

Molar mass of \(N_2\) is \(28.02\space g/mol\). Moles of \(N_2 = \frac{2.00\space g}{28.02\space g/mol} \approx 0.0714\space mol\).

Step2: Calculate moles of \(H_2\)

Molar mass of \(H_2\) is \(2.02\space g/mol\). Moles of \(H_2 = \frac{3.00\space g}{2.02\space g/mol} \approx 1.485\space mol\).

Step3: Determine limiting reactant

From reaction \(N_2 + 3H_2
ightarrow 2NH_3\), 1 mol \(N_2\) reacts with 3 mol \(H_2\). For \(0.0714\space mol\) \(N_2\), required \(H_2\) is \(0.0714 \times 3 = 0.2142\space mol\). We have \(1.485\space mol\) \(H_2\), so \(N_2\) is limiting.

Step4: Calculate moles of \(NH_3\)

From reaction, 1 mol \(N_2\) produces 2 mol \(NH_3\). Moles of \(NH_3 = 0.0714 \times 2 \approx 0.143\space mol\).

Answer:

0.143