Question

for the balanced equation, $2\ce{nh_3(g)} + 3\ce{cuo(s)} \
ightarrow \ce{n_2(g)} + 3\ce{cu(s)} + 3\ce{h_2o(l)}$, if you took 20.2mols of $\ce{nh_3}$ for an experiment, how many moles of $\ce{cuo}$ do you need for completing the reaction?.

1. 21.81mols cuo
2. 60.60mols cuo
3. 1.06 mols cuo
4. 30.3 mols cuo

Explanation:

Step1: Determine mole ratio

From the balanced equation \(2\text{NH}_3(\text{g}) + 3\text{CuO}(\text{s})
ightarrow \text{N}_2(\text{g}) + 3\text{Cu}(\text{s}) + 3\text{H}_2\text{O}(\text{l})\), the mole ratio of \(\text{NH}_3\) to \(\text{CuO}\) is \(2:3\).

Step2: Set up proportion

Let \(x\) be moles of \(\text{CuO}\). The proportion is \(\frac{2\ \text{mol}\ \text{NH}_3}{3\ \text{mol}\ \text{CuO}}=\frac{20.2\ \text{mol}\ \text{NH}_3}{x}\).

Step3: Solve for \(x\)

Cross - multiply: \(2x = 3\times20.2\). Then \(x=\frac{3\times20.2}{2}\). Calculate \(3\times20.2 = 60.6\), and \(\frac{60.6}{2}=30.3\) moles.

Answer:

1. 30.3 mols CuO