Calculate the mass of acetic acid that can be obtained using 500 g of technical calcium carbide, which contains 10.4% of impurities
m tech. (CaC2) = 500 g
ω approx. = 10.4%
m (CH3COOH) -?
1) CaC2 + 2H2O => C2H2 + Ca (OH) 2;
C2H2 + H2O => CH3CHO;
CH3CHO + Ag2O => CH3COOH + 2Ag;
2) ω (CaC2) = 100% – ω approx. = 100% – 10.4% = 89.6%;
3) m clean. (CaC2) = ω * m tech. / 100% = 89.6% * 500/100% = 448 g;
4) n (CaC2) = m pure. / M = 448/64 = 7 mol;
5) n (CH3COOH) = n (CH3CHO) = n (C2H2) = n (CaC2) = 7 mol;
6) m (CH3COOH) = n * M = 7 * 60 = 420 g.
Answer: The mass of CH3COOH is 420 g.
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