Given 7.75 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield? express your answer in grams to three significant figures.

Answer is: 9.69 grams of ethyl butyrate would be synthesized.

Chemical reaction: C₃H₇COOH + C₂H₅OH → C₃H₇COOC₂H₅ + H₂O.

m(C₃H₇COOH) = 7.35 g; mass of butanoic acid.

n(C₃H₇COOH) = m(C₃H₇COOH) ÷ M(C₃H₇COOH).

n(C₃H₇COOH) = 7.35 g ÷ 88.11 g/mol.

n(C₃H₇COOH) = 0.0834 mol; amount of butanoic acid.

From chemical reaction: n(C₃H₇COOH) : n(C₃H₇COOC₂H₅) = 1 : 1.

n(C₃H₇COOC₂H₅) = 0.0834 mol; amount of ethyl butyrate.

m(C₃H₇COOC₂H₅) = n(C₃H₇COOC₂H₅) · M(C₃H₇COOC₂H₅).

M(C₃H₇COOC₂H₅) = 6·Ar(C) + 2·Ar(O) + 12·Ar(H) · g/mol.

M(C₃H₇COOC₂H₅) = 6·12.01 + 2·15.99 + 12·1.01 · g/mol.

M(C₃H₇COOC₂H₅) = 116.16 g/mol; molar mass of ethyl butyrate.

m(C₃H₇COOC₂H₅) = 0.0834 mol · 116.16 g/mol.

m(C₃H₇COOC₂H₅) = 9.69 g.

indium, and thallium have three electrons i think