Answer
$CO$ is the limiting reactant.
Work Step by Step
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.
