A commercial method of manufacturing hydrogen involves the reaction of iron and steam. $$ 3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \stackrel{\Delta}{\longrightarrow} \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+4 \mathrm{H}_{2}(\mathrm{g}) $$ (a) How many grams of \(\mathrm{H}_{2}\) can be produced from \(42.7 \mathrm{g}\) Fe and an excess of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (steam)? (b) How many grams of \(\mathrm{H}_{2} \mathrm{O}\) are consumed in the conversion of \(63.5 \mathrm{g}\) Fe to \(\mathrm{Fe}_{3} \mathrm{O}_{4} ?\) (c) If \(14.8 \mathrm{g} \mathrm{H}_{2}\) is produced, how many grams of \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) must also be produced?

Step by step solution

01

Determine Molar Mass and Stoichiometric Ratios

From the periodic table, find the molar mass of the elements involved: Iron (Fe) = 55.845 g/mole, Hydrogen (H) = 1.008 g/mole, Oxygen (O) = 15.999 g/mole. Hence, determine the molar mass of water (H2O) and magnetite (Fe3O4). For H2O, it's \(2(1.008) g/mole + 15.999 g/mole = 18.015 g/mole\). For Fe3O4, it's \(3(55.845) g/mole + 4(15.999) g/mole = 231.533 g/mole\). The stoichiometric ratios are also important: 3 moles of Fe react with 4 moles of H2O to produce 1 mole of Fe3O4 and 4 moles of H2.

02

Calculate Quantity of Hydrogen (H2) Produced

To calculate the grams of H2 from 42.7g of Fe, first find the moles of Fe using \(moles = mass / molar mass\) which gives, \(42.7 g/55.845 g/mole = 0.765 moles of Fe\). Now using the stoichiometric ratio of 3 moles Fe : 4 moles H2 in the balanced equation, 0.765 moles of Fe will produce \(0.765*4/3 = 1.02 moles of H2\). Converting moles of H2 into mass gives \(1.02 moles*2(1.008 g/mole) = 2.052 g of H2\).

03

Calculate Quantity of Water (H2O) Consumed

Given 63.5 g of Fe, first find the moles of Fe with \(moles = mass / molar mass\) which gives, \(63.5 g / 55.845 g/mole = 1.137 moles of Fe\). Using the stoichiometric ratio of 3 moles Fe : 4 moles H2O in the balanced equation, 1.137 moles of Fe require \(1.137*4/3 = 1.516 moles of H2O\). Convert moles of H2O into mass for the final answer: \(1.516 moles * 18.015 g/mole = 27.31 g\).

04

Calculate Quantity of Fe3O4 Produced

Given 14.8 g of H2, first find the moles of H2 with \(moles = mass / molar mass\) which gives \(14.8 g / 2(1.008 g/mole) = 7.35 moles of H2\). Using the stoichiometric ratio of 4 moles H2 : 1 mole Fe3O4, 7.35 moles of H2 will produce \(7.35*1/4 = 1.8385 moles of Fe3O4\). Convert moles of Fe3O4 into mass for the final answer: \(1.8385 moles * 231.533 g/mole = 425.73 g\).

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