## Chemistry: Molecular Approach (4th Edition)

The theoretical yield of Ammonia is: $29.4$ kg.
** 1 kg = $10^3$ g $5.22 kg \times \frac{10^3g}{1kg} = 5.22 \times 10^3 g$ $31.5 kg \times \frac{10^3g}{1kg} = 31.5 \times 10^3 g$ 1. Calculate the number of moles of $H_2$: 1.008* 2 = 2.016g/mol $5.22 \times 10^3g \times \frac{1 mol}{ 2.016g} = 2.589 \times 10^3mol (H_2)$ According to the balanced equation: The ratio of $H_2$ to $NH_3$ is 3 to 2: $2.589 \times 10^3mol (H_2) \times \frac{ 2 mol(NH_3)}{ 3 mol (H_2)} = 1.726\times 10^3 mol (NH_3)$ --------- 3. Calculate the number of moles of $N_2$: 14.01* 2 = 28.02g/mol $31.5 \times 10^3g \times \frac{1 mol}{ 28.02g} = 1.124\times 10^3mol (N_2)$ According to the balanced equation: The ratio of $N_2$ to $NH_3$ is 1 to 2: $1.124 \times 10^3 mol (N_2) \times \frac{ 2 mol(NH_3)}{ 1 mol (N_2)} = 2.248 \times 10^3 mol (NH_3)$ --------------------------------------------------- Since $H_2$ made the least amount of $NH_3$, it is the limiting reactant, so we should follow its calculations with $1.726 \times 10^3$ moles of Ammonia: 2. Calculate the mass of $NH_3$: 14.01* 1 + 1.008* 3 = 17.03g/mol $1.726 \times 10^3 mol \times \frac{ 17.03 g}{ 1 mol} = 29.4 \times 10^3 g (NH_3) = 29.4 kg (NH_3)$