Calculate the mass of calcium carbide containing 25% impurities required to obtain 30 kg of acetic acid

Given:
ω approx. = 25%
m (CH3COOH) = 30 kg = 30,000 g

Find:
m (CaC2) -?

Solution:
1) CaC2 + 2H2O => Ca (OH) 2 + C2H2 ↑;
C2H2 + H2O = (Hg2 +) => CH3COH;
CH3COH = (CrO3, H2SO4) => CH3COOH;
2) Mr (CH3COOH) = Ar (C) * 2 + Ar (H) * 4 + Ar (O) * 2 = 12 * 2 + 1 * 4 + 16 * 2 = 60 g / mol;
Mr (CaC2) = Ar (Ca) + Ar (C) * 2 = 40 + 12 * 2 = 64 g / mol;
3) n (CH3COOH) = m (CH3COOH) / Mr (CH3COOH) = 30,000 / 60 = 500 mol;
4) n (CaC2) = n (C2H2) = n (CH3COH) = n (CH3COOH) = 500 mol;
5) m clean. (CaC2) = n (CaC2) * Mr (CaC2) = 500 * 64 = 32000 g;
6) ω (CaC2) = 100% – ω approx. = 100% – 25% = 75%;
7) m (CaC2) = m pure. (CaC2) * 100% / ω (CaC2) = 32000 * 100% / 75% = 42667 g.

Answer: The mass of CaC2 is 42667 g.