Take CO₂ = 44 g/mol. What mass of CO₂ is produced when 6.0 g of ethanoic acid (M = 60 g/mol) reacts with excess NaHCO₃?
Correct answer D. 4.4 g
Explanation
CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂, a 1:1 ratio. Moles of acid = 6.0/60 = 0.10 mol, so 0.10 mol CO₂ forms. Mass = 0.10 × 44 = 4.4 g.
Answer checked How we write and check questions
Discussion