Answer
$188g (H_2)$ can be produced.
Work Step by Step
As we determined on the last exercise, $CH_4$ is the limiting reactant, so the calculations are based on its mass:
1. Calculate the number of moles of $CH_4$:
12.01* 1 + 1.007* 4 = 16.05g/mol
$500g \times \frac{1 mol}{ 16.05g} = 31.17mol (CH_4)$
- The balanced reaction is:
$CH_4 + H_2O -- \gt CO + 3H_2$
The ratio of $CH_4$ to $H_2$ is 1 to 3:
$31.17 mol (CH_4) \times \frac{ 3 mol(H_2)}{ 1 mol (CH_4)} = 93.5mol (H_2)$
2. Calculate the mass of $H_2$:
1.007* 2 = 2.014g/mol
$93.5 mol \times \frac{ 2.014 g}{ 1 mol} = 188g (H_2)$
