дано
m(изв) = 120 g
W(CaCO3) = 20%
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m(CO2)-?
m(CaCO3) = 120 - (120*20% / 100%) =96 g
CaCO3--t-->CaO+CO2
M(CaCO3) = 100 g/mol
n(CaCO3) = m/M = 96/100 = 0.96 mol
n(CaCO3) = n(CO2) = 0.96 mol
M(CO2) = 44 g/mol
m(CO2) = n*M = 0.96 * 44 = 42.24 g
ответ 42.24 г