Question

When iron and steam react at high temperatures, the following reaction takes place. $$ 3 \mathrm{Fe}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{H}_{2}(g) $$ How much iron must react with excess steam to form \(897 \mathrm{~g}\) of \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) if the reaction yield is \(69 \%\) ?

Short answer

The mass of iron required to produce 897 g of iron oxide with a reaction yield of 69% is 939.55 g.

Step-by-step solution

Write the balanced chemical equation.

The given balanced chemical equation is: $$ 3 \mathrm{Fe}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{H}_{2}(g) $$

Calculate the molar mass of each compound.

To use stoichiometry, we will need to convert the given mass of iron oxide to moles. We begin by calculating the molar mass of each compound: Iron (Fe): \(55.845 \ \text{g/mol}\) Water (H_2O): \((2 \times 1.008) + 16 = 18.016 \ \text{g/mol}\) Iron oxide (Fe_3O_4): \((3 \times 55.845) + (4 \times 16) = 231.54 \ \text{g/mol}\) Hydrogen gas (H_2): \(2 \times 1.008 = 2.016 \ \text{g/mol}\)

Convert the mass of iron oxide to moles.

Using the molar mass of iron oxide, convert the given mass (897 g) to moles: $$ \text{moles of} \ Fe_{3}O_{4} = \frac{897 \ \text{g}}{231.54 \ \text{g/mol}} = 3.872 \ \text{moles} $$

Use stoichiometry to find moles of iron needed.

According to the balanced equation, 3 moles of iron react to produce 1 mole of iron oxide. We need to find how many moles of iron are needed: $$ \text{moles of} \ Fe = 3 \times \text{moles of} \ Fe_{3}O_{4} = 3 \times 3.872 \ \text{moles} = 11.616 \ \text{moles} $$

Convert moles of iron to mass in grams.

Now, convert the moles of iron needed to the mass in grams using the molar mass of iron: $$ \text{mass of} \ Fe = 11.616 \ \text{moles} \times 55.845 \ \text{g/mol} = 648.29 \ \text{g} $$

Consider the reaction yield.

Since the reaction has a yield of 69%, not all of the iron will react completely. Therefore, we need to adjust the mass of iron: $$ \text{adjusted mass of} \ Fe = \frac{\text{mass of Fe}}{\text{reaction yield}} = \frac{648.29 \ \text{g}}{0.69} = 939.55 \ \text{g} $$ In conclusion, 939.55 g of iron must react with excess steam to form 897 g of \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) if the reaction yield is 69%.