# Chapter 3 - Chemical Reactions - Questions for Review and Thought - Topical Questions - Page 149e: 70b

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.

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