Given 7.35g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield

 The reaction of Butanoic acid and excess ethanol occurs as shown in the equation;
 CH₃CH₂CH₂COOH + CH₃CH₂OH = CH₃CH₂CH₂COOCH₂CH₃ + H₂O
1 mole of butanoic acid contains 88g/mol
Thus, 7.35 g of butanoic acid contains; 
  = 7.35/88
  = 0.0835 moles
The mole ratio of butanoic acid and ethylbutyrate is 1:1
Thus moles of ethyl butyrate produced is 0.0835 moles
But 1 mole of ethyl butyrate contains 116 g/mol
Hence, the mass of ethyl butyrate will be
=0.0835 × 116 
=9.686 g

Answer is: 9,69 grams ethyl butyrate would be synthesized.
Chemical reaction: C₃H₇COOH + C₂H₅OH → C₃H₇COOC₂H₅ + H₂O.
m(C₃H₇COOH) = 7,35 g.
n(C₃H₇COOH) = m(C₃H₇COOH) ÷ M(C₃H₇COOH).
n(C₃H₇COOH) = 7,35 g ÷ 88,11 g/mol.
n(C₃H₇COOH) = 0,0834 mol.
From chemical reaction: n(C₃H₇COOH) : n(C₃H₇COOC₂H₅) = 1 : 1.
n(C₃H₇COOC₂H₅) = 0,0834 mol.
m(C₃H₇COOC₂H₅) = 0,0834 mol · 116,16 g/mol.
m(C₃H₇COOC₂H₅) = 9,69 g.