Stoichiometry with 50% yield for esterification

Steps:

• Molar mass of ethyl methanoate (HCOOCH₂CH₃): 74 g/mol; moles in 10 g: 10 / 74 ≈ 0.135 mol.
• Reaction is 1:1 (methanoic acid + ethanol → ester + water), so 0.135 mol alcohol reacts to form ester.
• At 50% conversion, total alcohol required: 0.135 mol / 0.5 = 0.27 mol.
• Molar mass of ethanol (CH₃CH₂OH): 46 g/mol; mass needed: 0.27 × 46 = 12.4 g.

Why C is correct:

• Matches mass of ethanol for 10 g ester via 1:1 stoichiometry adjusted for 50% yield in esterification.

Why the others are wrong:

• A: Mass for 100% yield (0.135 mol × 46 ≈ 6.2 g, halved incorrectly).
• B: Likely error in molar mass (e.g., using methanoic acid's 46 g/mol without yield adjustment).
• D: Excessive, possibly confusing ethanol with larger alcohol or ignoring stoichiometry.

Final answer: C