Answer
The mass of the excess reactant left after the reaction is equal to 1.3 g.
Work Step by Step
As we determined on the last exercise, $CO$ is the limiting reactant, so, the reaction calculations are based on the $CO$ mass:
1. Calculate the number of moles of $CO$:
12.01* 1 + 16* 1 = 28.01g/mol
$74.5g \times \frac{1 mol}{ 28.01g} = 2.66mol (CO)$
- The balanced reaction is:
$CO + 2H_2 -- \gt CH_3OH$
The ratio of $CO$ to $H_2$ is 1 to 2:
$2.66 mol (CO) \times \frac{ 2 mol(H_2)}{ 1 mol (CO)} = 5.32mol (H_2)$
2. Calculate the mass of $H_2$:
1.01* 2 = 2.02g/mol
$5.32 mol \times \frac{ 2.02 g}{ 1 mol} = 10.7g (H_2)$
We had, initially, 12.0 g of $H_2$; 10.7 g were consumed:
12.0 g - 10.7 g = 1.3 g.
This is the amount of $H_2$ that didn't react.
