дано
m техн ( CaC2) = 150 g
W(прим ) = 24%
η(C2H2) = 90%
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V пр (C2H2)-?
m(CaC2) = 150 -( 150 *24% / 100% ) = 114 g
CaC2+2H2O-->Ca(OH)2+C2H2
M(CaC2) = 64 g/mol
n(CaC2) = m/M = 114 / 64 = 1.78 mol
n(CaC2) = n(C2H2) = 1.78 mol
V теор (C2H2) = 1.78*22.4 = 39.872 L
V практ(C2H2) = V(C2H2) * η(C2H2) / 100% = 39.872 * 90% / 100% = 35.9 L
ответ 35.9 л