# Calculate the mass of calcium carbide obtained from 0.5 tons of limestone containing 15% impurities.

Given:
m (limestone) = 0.5 t = 500,000 g
ω approx. = 15%

To find:
m (CaC2) -?

Decision:
1) CaCO3 + 4C = (t-1500C) => CaC2 + 3CO ↑;
2) Mr (CaCO3) = Ar (Ca) + Ar (C) + Ar (O) * 3 = 40 + 12 + 16 * 3 = 100 g / mol;
Mr (CaC2) = Ar (Ca) + Ar (C) * 2 = 40 + 12 * 2 = 64 g / mol;
3) ω (CaCO3) = 100% – ω approx. = 100% – 15% = 85%;
4) m (CaCO3) = ω (CaCO3) * m (limestone) / 100% = 85% * 500000/100% = 425000 g;
5) n (CaCO3) = m (CaCO3) / Mr (CaCO3) = 425000/100 = 4250 mol;
6) n (CaC2) = n (CaCO3) = 4250 mol;
7) m (CaC2) = n (CaC2) * Mr (CaC2) = 4250 * 64 = 272000 g = 272 kg.

Answer: The mass of CaC2 is 272 kg.

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