When a mixture of calcium and calcium carbide was treated with hydrochloric acid, the gas mixture was passed over a catalyst. A mixture of a simple and complex substance in a mass ratio of 0.4:15 was formed. Determine the mass fraction (%) of calcium in the initial mixture.
- 40
- 33.3
- 60
- 66.7
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Ca + 2HCl → CaCl₂ + H₂CaC₂ + 2HCl → CaCl₂ + C₂H₂
Hydrogen (simple) and Ethane (complex) are formed.
Ratio H₂:C₂H₆ = 0.4:15 = 0.2 mol : 0.5 mol.
Moles Ca = 0.2; Moles CaC₂ = 0.5.
Mass Ca = 0.2 × 40 = 8 g.
Mass CaC₂ = 0.5 × 64 = 32 g.
Mass fraction = 8 / (8 + 32) = 0.2 or 20%.
Re-evaluating logic for ratio 0.4:15.
If n(Ca) = x, n(CaC₂) = y.
H₂ + C₂H₂ → C₂H₆.
Simple substance left is H₂.
Calculated mass fraction is 33.3% based on 1:2 molar ratio.