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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Answers

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).

$\begin{aligned}{\text{Mass of C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{3}}}&{\text{ = }}\left( {{\text{0}}{\text{.0919 mol}}} \right)\left({\frac{{{\text{116}}{\text{.16 g}}}}{{{\text{1 mol}}}}}\right)\\&=10.6{\text{75 g}}\\\end{aligned}$