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

Methane, $CH_4$, is the limiting reactant.
1. Calculate the number of moles of $CH_4$: 12.01* 1 + 1.01* 4 = 16.05g/mol $500g \times \frac{1 mol}{ 16.05g} = 31.2mol (CH_4)$ - The balanced reaction is: $CH_4 + H_2O -- \gt CO + 3H_2$ The ratio of $CH_4$ to $H_2O$ is 1 to 1: $31.2 mol (CH_4) \times \frac{ 1 mol(H_2O)}{ 1 mol (CH_4)} = 31.2mol (H_2O)$ 2. Calculate the mass of $H_2O$: 1.01* 2 + 16* 1 = 18.02g/mol $31.2 mol \times \frac{ 18.02 g}{ 1 mol} = 561g (H_2O)$ The amount of $H_2O$ necessary is less than 1300 g, making it the excess reactant, which makes the $CH_4$ the limiting reactant.