m(CaCO3) = 250 g
W(выхода CO2) = 90%
V(CO2) - ?
n(CaCO3) = 250 g / 100 g / mol = 2.5 mol
CaCO3 = CaO + CO2
n(CaCO3) = n(CO2) = 2.5 mol
V(CO2 теоретического) = 2.5 mol * 22.4 l / mol = 56 l
V(CO2 практического) = 56 l * 0.9 = 50.4 lОтвет: 50.4 l .
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m(CaCO3) = 250 g
W(выхода CO2) = 90%
V(CO2) - ?
n(CaCO3) = 250 g / 100 g / mol = 2.5 mol
CaCO3 = CaO + CO2
n(CaCO3) = n(CO2) = 2.5 mol
V(CO2 теоретического) = 2.5 mol * 22.4 l / mol = 56 l
V(CO2 практического) = 56 l * 0.9 = 50.4 l
Ответ: 50.4 l .