Answer
739 g of $H_2O$ didn't react.
Work Step by Step
As we determined in 71a, the limiting reactant is methane, so:
1. Calculate the number of moles of $CH_4$:
12.01* 1 + 1.008* 4 = 16.042g/mol
$500g \times \frac{1 mol}{ 16.042g} = 31.168mol (CH_4)$
- The balanced reaction is:
$CH_4 + H_2O -- \gt CO + 3H_2$
The ratio of $CH_4$ to $H_2O$ is 1 to 1:
$31.168 mol (CH_4) \times \frac{ 1 mol(H_2O)}{ 1 mol (CH_4)} = 31.168mol (H_2O)$
2. Calculate the mass of $H_2O$:
1.008* 2 + 16* 1 = 18.016g/mol
$31.168 mol \times \frac{ 18.016 g}{ 1 mol} = 561g (H_2O)$
561g of water reacted, and we had 1300 g initially:
1300 - 561 = 739 g
