1a) SOLUTION:

Step 1: Calculate the number of moles of N₂ formed by each reactant.

• Using NH₃

Based on the balanced chemical equation, 2 moles of NH₃ is stoichiometrically equivalent to 1 mole of N₂.

The molar mass of NH₃ is 17.031 g/mol.

[tex]\begin{aligned} \text{moles of} \: \text{N}_2 & = \text{9.05 g} \: \text{NH}_3 \times \frac{\text{1 mol} \: \text{NH}_3}{\text{17.031 g} \: \text{NH}_3} \times \frac{\text{1 mol} \: \text{N}_2}{\text{2 mol} \: \text{NH}_3} \\ & = \text{0.26569 mol} \end{aligned}[/tex]

• Using CuO

Based on the balanced chemical equation, 3 moles of CuO is stoichiometrically equivalent to 1 mole of N₂.

The molar mass of CuO is 79.545 g/mol.

[tex]\begin{aligned} \text{moles of} \: \text{N}_2 & = \text{45.2 g CuO} \times \frac{\text{1 mol CuO}}{\text{79.545 g CuO}} \times \frac{\text{1 mol} \: \text{N}_2}{\text{3 mol CuO}} \\ & = \text{0.18941 mol} \end{aligned}[/tex]

Step 3: Determine the limiting reagent and the excess reagent.

Since CuO produced less amount of N₂ than NH₃,

[tex]\boxed{\text{CuO is the limiting reagent and} \: \text{NH}_3 \: \text{is the excess reagent.}}[/tex]

1b) SOLUTION:

The theoretical yield of a reaction is the maximum mass of a product that can be formed. It is dictated by the limiting reactant. In this case, we will start at the number of moles of N₂ formed from the limiting reactant (CuO) which is equal to 0.18941 mol.

The molar mass of N₂ is 28.014 g/mol.

[tex]\begin{aligned} \text{theoretical yield} & = \text{0.18941 mol} \: \text{N}_2 \times \frac{\text{28.014 g} \: \text{N}_2}{\text{1 mol} \: \text{N}_2} \\ & = \text{5.3061 g} \\ & \approx \boxed{\text{5.31 g}} \end{aligned}[/tex]

Hence, the theoretical yield of the reaction is 5.31 g.

1c) SOLUTION:

[tex]\begin{aligned} \text{percent yield} & = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ & = \frac{\text{4.61 g}}{\text{5.3061 g}} \times 100\% \\ & = \boxed{86.9\%} \end{aligned}[/tex]

Hence, the percent yield of the reaction is 86.9%.

[tex]\\[/tex]

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