Chapter 11, Problem 79A

Interpretation Introduction

Interpretation:

The mass in grams for iron that is produced by the reaction of 25.0 moles of hematite with 30.0 moles of carbon monoxide needs to be determined.

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

Concept introduction:

The mole concept provides the relation between moles, mass and molar mass. The mathematical expression of mole concept can write as:

$\text{Mole =}\frac{\text{Mass}}{\text{Molar mass}}$

1 mole of any substance contains $\text{6}\text{.022 }\times {\text{ 10}}^{\text{23}}$ number of particles or atoms or molecules.

Answer to Problem 79A

$\text{1116 g Fe}$

Explanation of Solution

Given information:

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

Moles of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}$ = 25.0 moles

Moles of ${\text{CO}}_{\text{(g)}}$ = 30.0 moles

The given reaction:

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

1 mole of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}$ = 3 mole of ${\text{CO}}_{\text{(g)}}$

Determine the limiting reagent:

$\text{30}{\text{.0 moles CO}}_{\text{(g)}}\text{×}\frac{{\text{1 mole Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}}{{\text{3 mole CO}}_{\text{(g)}}}\text{=10}{\text{.0 moles Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}$

Since ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}$ is present in excess so ${\text{CO}}_{\text{(g)}}$ must be limiting reagent and will determine the amount of product form.

Calculate mass of Fe with the help of mole ratio:

$\text{30}\text{.0 moles CO ×}\frac{\text{2 mole Fe}}{\text{3 mole CO}}\text{×}\frac{\text{55}\text{.8 g Fe}}{\text{1 mole Fe}}\text{= 1116 g Fe}$

Conclusion

Mole ratio is used to calculate the mass and moles of product.