дано
m(CaC2) =1 t = 1000 kg
W(прим) = 30%
η(C2H2) = 80%
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Vпракт (C2H2)-?
mчист(CaC2)=1000 - (1000*30% / 100%) = 700 kg
CaC2+2H2O-->Ca(OH)2+C2H2
M(CaC2) = 64 kg/kmol
n(CaC2) = m/M = 700 / 64 = 10.94 kmol
n(CaC2) = n(C2H2) = 10.94 kmol
Vтеор (C2H2) = n*Vm = 10.94 * 22.4 = 245 m3
Vпракт(C2H2) = 245 * 80% / 100% = 196 m3
ответ 196 м3