Al₄C₃ + 12H₂O = 3CH₄ + 4Al(OH)₃

CaC₂ + 2H₂O = C₂H₂ + Ca(OH)₂

n (газов) = V / Vm = 4.48 / 22.4 = 0.2 моль

M (Al₄C₃) = 4Ar (Al) + 3Ar (C) = 27 * 4 + 12 * 3 = 144 г/моль

M (CaC₂) = Ar (Ca) + 2Ar (C) = 40 + 2 * 12 = 64 г/моль

Пусть x — количество Al₄C₃, y — количество CaC₂, тогда

144x + 64y = 10.4

3x + y = 0.2 ⇒ x = y = 0.05 моль

m (Al₄C₃) = n * M = 0.05 * 144 = 7.2 г

m (CaC₂) = n * M = 0.05 * 64 = 3.2 г

ω (Al₄C₃) = m (Al₄C₃) * 100% / m (смеси) = 7.2 г * 100% / 10.4 г = 69.23%

ω (CaC₂) = 100% - ω (Al₄C₃) = 100% - 69.23% = 30.77%

Ответ: ω (Al₄C₃) = 69.23%, ω (CaC₂) = 30.77%