дано
m(CaC2) = 26.6 kg
η(C2H2) = 85 %
-------------------
V практ (C2H2)-?
CaC2+2H2O-->Ca(OH)2+C2H2
M(CaC2) = 74 g/mol
n(CaC2) = m/M = 26.6 / 74 = 0.36 mol
n(CaC2) = n(C2H2) = 0.36 mol
Vтеор(C2H2) =n*Vm = 0.36 * 22.4 = 8.064 L
Vпракт(C2H2) = Vтеор(C2H2) * η(C2H2) / 100% = 8.064 * 85% / 100% = 6.85 L
ответ 6.85 л