Verified Answer :
Answer:
10.7 g of ethyl butyrate would be synthesized in 100 % conversion
Explanation:
1 mol of butanoic acid produces 1 mol of ethyl butyrate in 100% conversion.
Number of moles = (mass/molar mass)
Molar mass of butanoic acid = 88.11 g/mol
Molar mass of ethyl butyrate = 116.16 g/mol
So, 8.10 g of butanoic acid = \(\frac{8.10}{88.11}moles\)butanoic acid = 0.09193 moles of butanoic acid
So, in 100 % conversion, moles of ethyl butyrate would be produced = 0.09193 moles
So, in 100 % conversion, mass of ethyl butyrate would be produced = \((0.09193\times 116.16)g\)= 10.7 g
\(\boxed{{\text{10}}{\text{.675 g}}}\) of ethyl butyrate would be synthesized by 8.10 g of butanoic acid and excess ethanol.
Further Explanation:
Stoichiometry:
It is used to determine the amount of species present in the reaction by the relationship between reactants and products. It is used to determine the moles of a chemical species when moles of other chemical species present in the reaction is given.
Consider the general reaction,
\({\text{A}} + 2{\text{B}} \to 3{\text{C}}\)
Here,
A is the reactant.
B is the reactant.
C is the product.
One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.
The given reaction occurs as follows:
\({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{COOH}} + {\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}} \to {\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{COOC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_3} + {{\text{H}}_{\text{2}}}{\text{O}}\)
The formula to calculate the moles of \({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{COOH}}\) is as follows:
\({\text{Moles of C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{COOH}} = \frac{{{\text{Given mass of C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{COOH}}}}{{{\text{Molar mass of C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{COOH}}}}\) ...... (1)
Substitute 8.10 g for the given mass of \({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{COOH}}\)and 88.11 g/mol for the molar mass of \({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{COOH}}\)in equation (1).
\(\begin{aligned}{\text{Moles of C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{COOH}}&=\left({{\text{8}}{\text{.10 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{88}}{\text{.11 g}}}}}\right)\\&=0.091{\text{9 mol}}\\\end{aligned}\)
According to the stoichiometry of the reaction, one mole of \({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{COOH}}\) reacts with one mole of \({\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}\) to form one mole of \({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{COOC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_3}\)and one mole of \({{\text{H}}_2}{\text{O}}\). So the number of moles of \({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{COOC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_3}\) produced by 0.0919 moles of are also 0.0919 mol.
The formula to calculate the mass of \({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{COOC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_3}\) is as follows:
\({\text{Mass of C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{ = }}\left( {{\text{Moles of C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{3}}}}\)
\(\left( {{\text{Molar mass of C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{3}}}} \right)\) ...... (2)
Substitute 0.0919 mol for the moles of \({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{COOC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_3}\) and 116.16 g/mol for the molar mass of \({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{COOC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_3}\) in equation (2).
\(\begin{aligned}{\text{Mass of C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{3}}}&{\text{ = }}\left( {{\text{0}}{\text{.0919 mol}}} \right)\left({\frac{{{\text{116}}{\text{.16 g}}}}{{{\text{1 mol}}}}}\right)\\&=10.6{\text{75 g}}\\\end{aligned}\)
Since the yield is 100 %, so the amount of ethyl butyrate produced is 10.675 g.
Learn more:
1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride: https://brainly.com/question/3064603
2. Calculate the moles of ions in the solution: https://brainly.com/question/5950133
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Mole concept
Keywords: stoichiometry, ethyl butyrate, 0.0919 mol, one mole, 116.16 g/mol, 88.11 g/mol, 10.675 g.