Chapter 12, Problem 80A

(a)

Interpretation Introduction

Interpretation: The maximum mass of iron in grams is to be calculated.

Concept Introduction: The coefficients of a balanced chemical equation that are transformed into moles are used to calculate the mole ratio, which is a conversion factor.

When converting between a certain number of moles of one reactant or product and moles of another reactant or product, mole ratios are utilized.

Answer to Problem 80A

The maximum mass of iron in grams are $317.53\text{g}$ .

Explanation of Solution

Consider the chemical reaction:

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

Mass of iron oxide is $454\text{g}$ .

Molar mass of iron oxide is $159.688\text{g}$ .

Molar mass of iron is $55.845\text{g/mol}$ .

The moles of iron oxide reacted in the reaction are given as follows:

  $454\cancel{\text{g}{\text{Fe}}_2{\text{O}}_3}\times \frac{1\text{mol}{\text{Fe}}_2{\text{O}}_3}{159.688\cancel{\text{g}{\text{Fe}}_2{\text{O}}_3}}=2.843{\text{mol Fe}}_2{\text{O}}_3$

The moles of iron in iron oxide is given as follows:

  $2.843\cancel{{\text{mol Fe}}_2{\text{O}}_3}\times \frac{2\text{mol}\text{Fe}}{1\cancel{\text{mol}{\text{Fe}}_2{\text{O}}_3}}=5.686\text{mol Fe}$

The mass of iron in iron oxide is given as follows:

  $5.686\cancel{\text{mol Fe}}\times \frac{55.845\text{g}}{1\cancel{\text{mol}\text{Fe}}}=317.53\text{g}$

The mass of iron is $317.53\text{g}$ .

(b)

Interpretation Introduction

Interpretation: The mass of $\text{CO}$ required to reduce the iron (III) oxide to iron metal is to be calculated.

Concept Introduction: The coefficients of a balanced chemical equation that are transformed into moles are used to calculate the mole ratio, which is a conversion factor.

When converting between a certain number of moles of one reactant or product and moles of another reactant or product, mole ratios are utilized.

Answer to Problem 80A

The mass of $\text{CO}$ required to reduce the iron (III) oxide to iron metal is $238.8\text{g}$ .

Explanation of Solution

Consider the chemical reaction:

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

Mass of iron oxide is $454\text{g}$ .

Molar mass of iron oxide is $159.688\text{g}$ .

Molar mass of iron is $55.845\text{g/mol}$ .

The moles of iron oxide reacted in the reaction are given as follows:

  $454\cancel{\text{g}{\text{Fe}}_2{\text{O}}_3}\times \frac{1\text{mol}{\text{Fe}}_2{\text{O}}_3}{159.688\cancel{\text{g}{\text{Fe}}_2{\text{O}}_3}}=2.843{\text{mol Fe}}_2{\text{O}}_3$

The moles of $\text{CO}$

reacted with iron oxide are given as follows:

  $2.843\cancel{{\text{mol Fe}}_2{\text{O}}_3}\times \frac{3\text{mol}\text{CO}}{1\cancel{\text{mol}{\text{Fe}}_2{\text{O}}_3}}=8.52\text{mol CO}$

Molar mass of ye $\text{CO}$ is $28.01\text{g/mol}$ .

The mass of $\text{CO}$

reacted with iron oxide is given as follows:

  $8.52\cancel{\text{mol CO}}\times \frac{28.01\text{g}}{1\cancel{\text{mol}\text{CO}}}=238.8\text{g}$

The mass of 1y $\text{CO}$ is $238.8\text{g}$ .