Ответ:
дано
m(ppa CaCO3) = 300 g
W(CaCO3) = 0.8
+HCL
---------------------
V(CO2)=?
m(CaCO3) = 300 * 0.8 = 240 g
M(CaCO3) = 100g / mol
n(CaCO3) = m / M = 240 / 100 = 2.4 mol
CaCO3+2HCL-->CaCL2+H2O+CO2
n(CaCO3) = n(CO2) = 2.4 mol
V(CO2) = n*Vm = 2.4 * 22.4 = 53.76 L
ответ 53.76 л
Объяснение:
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Answers & Comments
Ответ:
дано
m(ppa CaCO3) = 300 g
W(CaCO3) = 0.8
+HCL
---------------------
V(CO2)=?
m(CaCO3) = 300 * 0.8 = 240 g
M(CaCO3) = 100g / mol
n(CaCO3) = m / M = 240 / 100 = 2.4 mol
CaCO3+2HCL-->CaCL2+H2O+CO2
n(CaCO3) = n(CO2) = 2.4 mol
V(CO2) = n*Vm = 2.4 * 22.4 = 53.76 L
ответ 53.76 л
Объяснение: