дано
m техн (CaCO3) = 10 g
W прим (CaCO3) = 10%
m(HCL) = 20 g
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V(CO2)-?
mчист(CaCO3) = mтехн(CaCO3) - (mтехн(CaCO3) * W(CaCO3) / 100%)
m чист (CaCO3)= 10 - (10*10%/100%) = 9 g
M(CaCO3) = 100 g/mol
n(CaCO3) = m/M = 9 / 100 = 0.09 mol
M(HCL) = 36.5 g/mol
n(HCL) = m(HCL) / M(HCL) = 20 / 36.5 = 0.55 mol
n(CaCO3)
CaCO3+2HCL-->CaCL2+H2O+CO2↑
n(CaCO3) = n(CO2) = 0.09 mol
V(CO2) = n(CO2) * Vm = 0.09 * 22.4 = 2.016 L
ответ 2.016 л