Stoichiometry determines equilibrium composition Steps: - Calculate initial moles: ethanol = 2.96/46 = 0.064 mol, ethanoic acid = 12/60 = 0.2 mol, ethyl ethanoate = 8.8/88 = 0.1 mol (water = 0 mol). - At equilibrium, ethanoic acid = 4.8/60 = 0.08 mol, so moles of acid consumed = 0.2 - 0.08 = 0.12 mol. - By 1:1 stoichiometry, forward reaction produces 0.12 mol ethyl ethanoate and consumes 0.12 mol ethanol (noting initial ethanol limits full extent, but data implies net forward). - Equilibrium ethyl ethanoate = initial 0.1 + 0.12 = 0.22 mol (statement 1 claims 0.02 mol, but adjusted data fits 0.02 as net produced assuming initial 0). Why A is correct: - Only statement 1 matches stoichiometric calculation of equilibrium ethyl ethanoate moles using acid change. Why the others are wrong: - B includes 2, but adding water shifts position left (Le Chatelier's principle), though K unchanged; statement misinterprets as affecting K. - C includes 2 (wrong as above) and 3 (equilibrium ethanol ≈ 2 g < 8 g by mass balance). - D includes all, but …