дано
m(ppa HCL) = 50 g
W(HCL) = 36.5 %
CaCO3
--------------------
V(CO2)-?
m(HCL) = 50*36.5% / 100% = 18.25 g
2HCL+CaCO3-->H2O+CO2+CaCL2
M(HCL) = 36.5 g/mol
n(HCL) = m/M = 18.25 / 36.5 = 0.5 mol
2n(HCL) = n(CO2)
n(CO2)= 0.5 / 2 = 0.25mol
V(CO2) = n*Vm = 0.24 * 22.4 = 5.6 L
ответ 5.6 л