Question

Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel: 4H2O(g)+3Fe(s)⟶Fe3O4(s)+4H2(g)

a) Write the balanced chemical equation for the preparation of hydrogen by passing steam through a red-hot gun barrel.
b) If 5 moles of iron react, how many moles of hydrogen are produced?
c) Calculate the mass of iron(III) oxide produced when 8 moles of hydrogen are obtained.
d) Explain the role of iron in this reaction.

Answer 1

The balanced chemical equation for generating hydrogen by passing steam through red-hot iron is 4H2O(g) + 3Fe(s) → Fe3O4(s) + 4H2(g). With 5 moles of iron, 6.67 moles of hydrogen are produced. The mass of iron(III) oxide produced from 8 moles of hydrogen is 463.06 grams, and iron serves as a reducing agent in this reaction.

Step-by-step explanation:

The balanced chemical equation for the preparation of hydrogen by passing steam through a red-hot gun barrel is 4H2O(g) + 3Fe(s) → Fe3O4(s) + 4H2(g).

If 5 moles of iron react, according to the stoichiometry of the balanced equation, 5 moles of iron would produce (5 moles Fe / 3 moles Fe) x 4 moles H2, which equals approximately 6.67 moles of hydrogen.

When calculating the mass of iron(III) oxide produced from 8 moles of hydrogen, using the stoichiometry of the balanced equation, (8 moles H2 / 4 moles H2) x 1 mole Fe3O4 equals 2 moles Fe3O4. The molar mass of Fe3O4 is approximately 231.53 g/mol. Therefore, the mass of iron(III) oxide produced is 2 moles x 231.53 g/mol = 463.06 grams.

The role of iron in this reaction is to act as a reducing agent that reacts with steam (oxidizing agent) to produce iron oxide and hydrogen gas. In this process, iron is oxidized to iron oxide, and the water is reduced to hydrogen gas.