ammonium sulfate, an important fertilizer,\ncan be prepared by the reaction of\nammonia with sulfuric…

ammonium sulfate, an important fertilizer,\ncan be prepared by the reaction of\nammonia with sulfuric acid.\n$2\\text{nh}_3(g) + \\text{h}_2\\text{so}_4(aq) \\rightarrow (\\text{nh}_4)_2\\text{so}_4(aq)$\nhow many moles of $\\text{nh}_3$ are needed to react\nwith 12.0 l of 0.500 m $\\text{h}_2\\text{so}_4$\n? mol $\\text{nh}_3$\ndo not round your answer.

ammonium sulfate, an important fertilizer,\ncan be prepared by the reaction of\nammonia with sulfuric acid.\n$2\\text{nh}_3(g) + \\text{h}_2\\text{so}_4(aq) \\rightarrow (\\text{nh}_4)_2\\text{so}_4(aq)$\nhow many moles of $\\text{nh}_3$ are needed to react\nwith 12.0 l of 0.500 m $\\text{h}_2\\text{so}_4$\n? mol $\\text{nh}_3$\ndo not round your answer.

Answer

Explanation:

Step1: Calculate moles of $\text{H}_2\text{SO}_4$

Use $n = M \times V$, where $M$ is molarity, $V$ is volume. $n_{\text{H}_2\text{SO}_4} = 0.500\ \text{mol/L} \times 12.0\ \text{L} = 6.00\ \text{mol}$

Step2: Relate moles via reaction ratio

From the balanced equation, 2 mol $\text{NH}_3$ reacts with 1 mol $\text{H}_2\text{SO}4$. $n{\text{NH}_3} = 6.00\ \text{mol}\ \text{H}_2\text{SO}_4 \times \frac{2\ \text{mol}\ \text{NH}_3}{1\ \text{mol}\ \text{H}_2\text{SO}_4}$

Step3: Compute final moles of $\text{NH}_3$

$n_{\text{NH}_3} = 6.00 \times 2 = 12.0\ \text{mol}$

Answer:

12.0 mol $\text{NH}_3$