Chapter 11, Problem 63A

Interpretation Introduction

Interpretation:

The mass of ethyl butanoate produces from 4.50 moles of ethanol needs to be deermined according to given balance chemical equation: ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l) }}{\text{ + C}}_{\text{3}}{\text{H}}_{\text{7}}{\text{COOH}}_{\text{(l) }}\stackrel{}{\to }{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5(l)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

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 63A

$\text{522 g ethyl butanoate }$

Explanation of Solution

The reaction of ethanol with butanoic acid forms ethyl butanoate and water.

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l) }}{\text{ + C}}_{\text{3}}{\text{H}}_{\text{7}}{\text{COOH}}_{\text{(l) }}\stackrel{}{\to }{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5(l)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

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

Moles of ethanol = 4.50 moles

1 moles of ethanol = 1 mole of ethyl butanoate

Molar mass of ethyl butanoate = 116 g/mol

Calculate mass of ethyl butanoate with the help of mole ratio:

$\text{4}\text{.50 moles ethanol ×}\frac{\text{1 mole ethyl butanoate }}{\text{1 mole ethanol }}\text{×}\frac{\text{116 g ethyl butanoate }}{\text{1 mole ethyl butanoate }}\text{= 522 g ethyl butanoate }$

Conclusion

Thus, $\text{522 g ethyl butanoate }$ is formed.