Chapter 11, Problem 69A

Interpretation Introduction

Interpretation:

The mass of ${\text{CO}}_{\text{2}}$ produces from 100.0 g of ethanol for the given reaction needs to be determined.

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 69A

Mass = $\text{ 95}{\text{.6 g CO}}_{\text{2 }}$

Explanation of Solution

Given information:

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l)}}{\text{ + O}}_{\text{2}}{}_{\text{(g) }}\stackrel{}{\to }{\text{ CO}}_{\text{2(g) }}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

Mass of ethanol = 100.0 g

The given reaction:

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l)}}{\text{ + O}}_{\text{2}}{}_{\text{(g) }}\stackrel{}{\to }{\text{ CO}}_{\text{2(g) }}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

Mass of ethanol = 100.0 g

Molar mass of ethanol = 46.0 g/mol

Molar mass of ${\text{CO}}_{\text{2}}$ = 44.0 g/mol

The mole ratio relates the moles of reactant with other reactant and product molecules in the balance chemical equation.

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l)}}{\text{ + O}}_{\text{2}}{}_{\text{(g) }}\stackrel{}{\to }{\text{ CO}}_{\text{2(g) }}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

1 mole of ${\text{CO}}_{\text{2}}$ = 1 mole of ethanol

Calculate mass of ${\text{CO}}_{\text{2}}$ with the help of mole ratio:

$\text{100}{\text{.0 g C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH ×}\frac{{\text{1 mole C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}{\text{46}{\text{.0 g C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}\text{×}\frac{{\text{1 mole CO}}_{\text{2 }}}{{\text{1 mole C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH }}\text{×}\frac{\text{44}{\text{.0 g CO}}_{\text{2 }}}{{\text{1 mole CO}}_{\text{2 }}}\text{= 95}{\text{.6 g CO}}_{\text{2 }}$

Conclusion

Mole ratio is used to calculate the mass of product.