# Question #dc3c2

May 6, 2016

First write the balanced oxidation-reduction reaction, then determine the number of moles of iron, then moles of ${H}_{2}$, then calculate the volume of gas using the ideal gas law , and finally multiply by 0.75 for the yield. (whew!)

#### Explanation:

$F e + 2 {H}^{+} \to F {e}^{2 +} + {H}_{2}$

$\frac{11.2 g}{55.845 \frac{g}{m o l}} = 0.201$ mol Fe = 0.201 mol ${H}_{2}$

STP = standard temperature and pressure:
1 atm pressure and 273.15 K

Using the ideal gas law, where R = 0.08206 L-atm/(mol-K):

$V = \frac{n R T}{P} = \frac{0.201 \cdot 0.08206 \cdot 273}{1} = 4.50 L$

75% yield gives $V = 4.50 L \times 0.75 = 3.38 L$