Gas formation from reaction #3

A mixture of iron (II) sulfide contains 93% of iron sulfide and 3% of metallic iron. Compute the volume (at standard condition) of gas that is formed when 10 grams of the mixture react with acid. The two reactions that occur are:
$$FeS+H^+ \rightarrow Fe^{2+}+H_2S$$
$$Fe+H^+ \rightarrow Fe^{2+}+H_2$$

SOLUTION:
The first thing to do is balance the two reactions:
$$FeS+{\color{red} 2}H^+ \rightarrow Fe^{2+}+H_2S$$
$$Fe+{\color{red} 2}H^+ \rightarrow Fe^{2+}+H_2$$

Since the 10 grams of the mixture are 97% of FeS and 3% for Fe, we can calculate the amount in grams of each component, and, knowing the molecular weight, calculate the number of moles, as follows:

$$97\%(10gr)=9.7gr(FeS)$$
$$3\%(10gr)=0.3gr(Fe)$$

Number of moles:
$$n_{FeS}=\frac{m_{FeS}}{MW_{FeS}}=\frac{9.7gr}{(55.85+32.07)gr/mol}=0.11mol$$

$$n_{Fe}=\frac{m_{Fe}}{MW_{Fe}}=\frac{0.3gr}{55.85gr/mol}=0.005mol$$

Since both the hydrogen sulfide and pure hydrogen are gas, we must calculate the number of moles of both species. Knowing the stoichiometry of the two reactions, we deduce that:

1) for each mole of Fes that reacts, it forms one mole of $H_2S$:

$$n_{H_2S}=n_{FeS}=0.11mol$$

Since a mole of gas is 22.4 liters, we have that the volume of H_2S that is formed is equal to:
$$V_{H_2S}=n_{H_2S}=0.11mol \cdot 22.4L/mol=2.464L$$

2) for each mole of iron metal that reacts, it forms a mole of hydrogen gas:

$$n_{H_2}=n_{Fe}=0.005mol$$
$$V_{H_2}=0.005mol \cdot 22.4L/mol=0.112L$$

The total volume of gas that is formed is then:

$$V_{tot}=V_{H_2S}+V_{H_2}=2.576L$$

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PROBLEM 20:
how many liters of chlorine are required to transform 50.5 grams of $PCl_3$ in $PCl_5$?