Дано
m(CaCO3) = 40 g
W(прим) = 20%
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V(CO2)-?
n(CO2)-?
M(CO2)-?
m(CO2)-?
m чист (CaCO3) = 40-40*20% / 100% = 32 g
CaCO3-->CaO+CO2
M(CaCO3) = 100 g/mol
n(CaCO3) = m/M = 32 / 100 = 0.32 mol
n(CaCO3) = n(CO2)= 0.32 mol
V(CO2) = n*Vm = 0.32*22.4 = 7.168 L
M(CO2) = 44 g/mol
m(CO2) = n*M = 0.32 * 44 = 14.08 g
ответ :V(CO2) = 7.168 л , М(СО2) = 44 г/моль , Mr(CO2) = 44 . n(CO2) = 0.32 mol . m(CO2) = 14.08 g