дан
m(CaO) = 140 kg
m практ (Ca(OH)2) = 182 kg
η(Ca(OH)2)-?
CaO+2H2O--->Ca(OH)2
M(CaO) = 56 kg/kmol
n(CaO) = m/M = 140 / 56 = 2.5 kmol
n(CaO) = n(Ca(OH)2) = 2.5 kmol
M(Ca(OH)2) = 74 kg/kmol
m теор(Ca(OH)2) = n*M = 2.5 * 74 = 185 kg
η(Ca(OH)2) = m практ (Ca(OH)2) /mтеор (Ca(OH)2) * 100% = 182 / 185 * 100% = 98.4%
ответ 98.4%
CaO + H2O = Ca (OH)2
56г/моль 74г/моль
185-----100%
182------У%
У%=98,4%
ответ: 98,4
дан
m(CaO) = 140 kg
m практ (Ca(OH)2) = 182 kg
η(Ca(OH)2)-?
CaO+2H2O--->Ca(OH)2
M(CaO) = 56 kg/kmol
n(CaO) = m/M = 140 / 56 = 2.5 kmol
n(CaO) = n(Ca(OH)2) = 2.5 kmol
M(Ca(OH)2) = 74 kg/kmol
m теор(Ca(OH)2) = n*M = 2.5 * 74 = 185 kg
η(Ca(OH)2) = m практ (Ca(OH)2) /mтеор (Ca(OH)2) * 100% = 182 / 185 * 100% = 98.4%
ответ 98.4%