Question

1.30g H2 is allowed to react with 10.4g N2 producing 2.74g NH3. Calculate the theoretical yield for the reaction and the percentage yield of the reaction

Answer 1

Answer on Question #39969 - Chemistry – Inorganic Chemistry

Question

1.30g H₂ is allowed to react with 10.4g N₂ producing 2.74g NH₃. Calculate the theoretical yield for the reaction and the percentage yield of the reaction.

Answer:

A balanced equation for the reaction:

$$3H_2 + N_2 = 2NH_3$$

Number of moles equals:

$$n = \frac{m}{M}$$

m – Mass of the substance, g.

M – Molar mass of the substance, g/mol.

Molar masses of the reactants equal:

$$M(H_2) = 2.016 \text{ g/mol}, \quad M(N_2) = 28.014 \text{ g/mol}$$

Number of moles of the reactants:

$$n(H_2) = \frac{m(H_2)}{M(H_2)} = \frac{1.30}{2.016} = 0.645 \text{ moles}$$ $$n(N_2) = \frac{m(N_2)}{M(N_2)} = \frac{10.4}{28.014} = 0.371 \text{ moles}$$

Then we make a proportion:

$$3 \text{ moles of } H_2 \text{ react with } 1 \text{ moles of } N_2$$ $$0.645 \text{ moles of } H_2 - x \text{ moles of } N_2$$ $$x = \frac{0.645 \cdot 1}{3} = 0.215 \text{ moles of } N_2 \text{ will react with } 0.645 \text{ moles of } H_2$$

We have 0.371 moles of nitrogen, therefore it is in excess. If it is in excess then the hydrogen is the limiting reactant.

We need to make another proportion to calculate the theoretical yield for the reaction:

$$3 \text{ moles of } H_2 \text{ produce } 2 \text{ moles of } NH_3$$ $$0.645 \text{ moles of } H_2 - x \text{ moles of } NH_3$$ $$x = \frac{0.645 \cdot 2}{3} = 0.43 \text{ moles of } NH_3 \text{ could be produced}$$

The theoretical yield for the reaction equals:

$$m _ {t} \left(N H _ {3}\right) = n \left(N H _ {3}\right) \cdot M \left(N H _ {3}\right) = 0.43 \cdot 15.015 = 6.46 \text{ g}$$

The percentage yield of the reaction equals:

$$\% \text{yield} = \frac{m_p(NH_3)}{m_t(NH_3)} \times 100\% = \frac{2.74}{6.46} \times 100\% = 42.4\%$$

Answer: Theoretical yield equals 6.46 g of NH₃, percentage yield equals 42.4%.