## Chemistry: The Molecular Science (5th Edition)

Mass of $N_2O$ produced: 5.50g Mass of water produced: 4.51g
1. Calculate the number of moles of $NH_4NO_3$: 14.01* 1 + 1.01* 4 + 14.01* 1 + 16* 3 = 80.06g/mol $10g \times \frac{1 mol}{ 80.06g} = 0.125mol (NH_4NO_3)$ As we determined in the last exercise, this is the balanced reaction: $NH_4NO_3 \lt -- \gt N_2O + 2H_2O$ The ratio of $NH_4NO_3$ to $N_2O$ is 1 to 1: $0.125 mol (NH_4NO_3) \times \frac{ 1 mol(N_2O)}{ 1 mol (NH_4NO_3)} = 0.125mol (N_2O)$ 2. Calculate the mass of $N_2O$: 14.01* 2 + 16* 1 = 44.02g/mol $0.125 mol \times \frac{ 44.02 g}{ 1 mol} = 5.50g (N_2O)$ --------------- The ratio of $NH_4NO_3$ to $H_2O$ is 1 to 2: $0.125 mol (NH_4NO_3) \times \frac{ 2 mol(H_2O)}{ 1 mol (NH_4NO_3)} = 0.25mol (H_2O)$ 2. Calculate the mass of H2O: 1.01* 2 + 16* 1 = 18.02g/mol $0.250 mol \times \frac{ 18.02 g}{ 1 mol} = 4.51g (H_2O)$