Chemistry, 14.10.2019 15:30 gioc3948

Given 8.10g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield?

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Answer from: rsanchez1226

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

Answer from: ununoctrium4414

$\boxed{{\text{10}}{\text{.675 g}}}$ 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:

2. Calculate the moles of ions in the solution:

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.

Answer from: Quest

The correct answer would be 1. precipitation

Answer from: Quest

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.

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Given 8.10g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesiz...

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