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

$CO$ is the limiting reactant.
1. Calculate the number of moles of $H_2$: 1.01* 2 = 2.02g/mol $12g \times \frac{1 mol}{ 2.02g} = 5.94mol (H_2)$ - The balanced reaction is: $CO + 2H_2 -- \gt CH_3OH$ The ratio of $H_2$ to $CO$ is 2 to 1: $5.94 mol (H_2) \times \frac{ 1 mol(CO)}{ 2 mol (H_2)} = 2.97mol (CO)$ 2. Calculate the mass of $CO$: 12.01* 1 + 16* 1 = 28.01g/mol $2.97 mol \times \frac{ 28.01 g}{ 1 mol} = 83.2g (CO)$ The necessary mass of $CO$ is higher than 74.5g Therefore, $CO$ is the limiting reactant.