Chapter 7, Problem 42A

Interpretation Introduction

Interpretation:

The following equations need to be balanced:

  $\begin{array}{l}{\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}{\text{(s) + H}}_{\text{2}}\text{(g)}\to {\text{Fe(s) + H}}_{\text{2}}\text{O(g) ------(1)}\\ {\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\text{(s) + CO(g)}\to {\text{Fe(s) + CO}}_{\text{2}}\text{(g) ------(2)}\end{array}$

Concept Introduction:

• A chemical reaction is expressed as a chemical equation having reactants and products on left and right side of the reaction arrow respectively.

  $\text{Reactants }\to \text{ Products}$

• A balanced chemical equation is one in which the number of atoms of various elements on the reactants side is equal to that on the product side. It depicts the stoichiometry of a given reaction.
• The equations are balanced in two steps:

1) Balance the atoms by matching the number of atoms of each type on the reactants and product side.

2) Adjust the coefficients to the smallest whole number. A coefficient of '1' is implied and therefore not explicitly written.

Answer to Problem 42A

  ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}{\text{(s) + 4H}}_{\text{2}}\text{(g)}\to 3{\text{Fe(s) + 4H}}_{\text{2}}\text{O(g) }$

  ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\text{(s) + 4CO(g)}\to 3{\text{Fe(s) + 4CO}}_{\text{2}}\text{(g) }$

Explanation of Solution

1) The first equation involves the reaction of solid iron oxide with hydrogen gas to form elemental iron and water vapor. The reactants and products are:

Reactants:

  $\begin{array}{l}{\text{Iron oxide, solid - Fe}}_{\text{3}}{\text{O}}_{\text{4}}\text{ (s)}\\ {\text{Hydrogen, gas - H}}_{\text{2}}(\text{g})\end{array}$

Products:

  $\begin{array}{l}{\text{Water vapor - H}}_2\text{O (g)}\\ \text{Iron, solid - Fe (s)}\end{array}$

Unbalanced equation:

  ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}{\text{(s) + H}}_{\text{2}}\text{(g)}\to {\text{Fe(s) + H}}_{\text{2}}\text{O(g) }$

The above equation is not balanced since the number of atoms of Fe and O on the reactant side is3 and 4 respectively while it is one each on the products side. In order to balance the above equation, multiply both H₂ and H₂O by 4 and Fe by 3 to get:

  ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}{\text{(s) + 4H}}_{\text{2}}\text{(g)}\to 3{\text{Fe(s) + 4H}}_{\text{2}}\text{O(g) }$

2) The second equation involves the reaction of solid iron oxide with carbon monoxide gas to form elemental iron and carbon dioxide. The reactants and products are:

Reactants:

  $\begin{array}{l}{\text{Iron oxide, solid - Fe}}_{\text{3}}{\text{O}}_{\text{4}}\text{ (s)}\\ \text{Carbon monoxide, gas - CO}(\text{g})\end{array}$

Products:

  $\begin{array}{l}{\text{Carbon dioxide - CO}}_2\text{(g)}\\ \text{Iron, solid - Fe (s)}\end{array}$

Unbalanced equation:

  ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\text{(s) + CO(g)}\to {\text{Fe(s) + CO}}_{\text{2}}\text{(g) }$

The above equation is not balanced since the number of atoms of Fe and O on the reactant side is 3 and 5 respectively while on the products side it is 1 and 2 respectively. In order to balance the above equation, multiply both CO and CO₂ by 4 and Fe by 3 to get:

  ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\text{(s) + 4CO(g)}\to 3{\text{Fe(s) + 4CO}}_{\text{2}}\text{(g) }$

Conclusion

Therefore, the balanced equations are:

  ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}{\text{(s) + 4H}}_{\text{2}}\text{(g)}\to 3{\text{Fe(s) + 4H}}_{\text{2}}\text{O(g) }$

  ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\text{(s) + 4CO(g)}\to 3{\text{Fe(s) + 4CO}}_{\text{2}}\text{(g) }$