A restaurant owner went to clean his oven using a solution that contained 150 g of sodium hydroxide (NaOH, 40.0 g/mol) . He decided to neutralize the excess solution with an amount of vinegar containing 300 g of acetic acid (CH₃COOH, 60.1 g/mol) . The products are sodium acetate (82.0 g/mol) and water (18.0 g/mol) . Identify the excess reactant and the amount that remains unreacted. The neutralization reaction equation is
NaOH(aq) + CH₃COOH(aq) $\rightarrow$ NaOOCCH₃(aq) + H₂O(l)

Question

A restaurant owner went to clean his oven using a solution that contained 150 g of sodium hydroxide (NaOH, 40.0 g/mol) . He decided to neutralize the excess solution with an amount of vinegar containing 300 g of acetic acid (CH₃COOH, 60.1 g/mol) . The products are sodium acetate (82.0 g/mol) and water (18.0 g/mol) . Identify the excess reactant and the amount that remains unreacted. The neutralization reaction equation is
NaOH(aq) + CH₃COOH(aq)

\rightarrow

NaOOCCH₃(aq) + H₂O(l)

Quizplus · Accepted Answer

The moles of NaOH are 150 g / 40.0 g/mol = 3.75 mol. The moles of acetic acid are 300 g / 60.1 g/mol = 4.99 mol. The reaction is 1:1, so NaOH is the limiting reactant. After reacting 3.75 mol of acetic acid, 4.99 - 3.75 = 1.24 mol of acetic acid remain, which is 1.24 mol * 60.1 g/mol = 74.6 g.

A Restaurant Owner Went to Clean His Oven Using a Solution