Question

2nh₃(g) + 3cuo(s) → n₂(g) + 3cu(s)+ 3h₂o(l).
therefore, first be mindful of stoichiometric ratio between nh₃ and cuo reactants and cu product. suppose, 5moles of nh₃ and 10 moles of cuo were taken for an experiment, first determine which one is the limiting reactant by comparing to the stoichiometric required mols. then, based on the limiting reactant, determine the theoretical yield (in moles) for copper that will be formed. (no need to convert to grams in any step).

○ 1) 5moles of cu

○ 2) 15 moles of cu

○ 3) 7.5moles of cu

○ 4) 1 mole of cu

Explanation:

Step1: Find stoichiometric ratio

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

Step2: Calculate required moles for \(\text{NH}_3\)

If we use 5 moles of \(\text{NH}_3\), moles of \(\text{CuO}\) required \(=\frac{3}{2} \times 5 = 7.5\) moles. We have 10 moles of \(\text{CuO}\), which is more than 7.5, so \(\text{NH}_3\) is not limiting.

Step3: Calculate required moles for \(\text{CuO}\)

If we use 10 moles of \(\text{CuO}\), moles of \(\text{NH}_3\) required \(=\frac{2}{3} \times 10 \approx 6.67\) moles. We have 5 moles of \(\text{NH}_3\), which is less than 6.67, so \(\text{NH}_3\) is the limiting reactant.

Step4: Calculate moles of \(\text{Cu}\) from \(\text{NH}_3\)

From the reaction, 2 moles of \(\text{NH}_3\) produce 3 moles of \(\text{Cu}\). Moles of \(\text{Cu}\) from 5 moles of \(\text{NH}_3=\frac{3}{2} \times 5 = 7.5\) moles.

Answer:

1. 7.5moles of Cu