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MaximLopnul
@MaximLopnul
July 2022
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акой V газа выдел. при взаимодействии 150г. 5% р-ра серной кислоты с раствором карбоната натрия?
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nikolaygr
Verified answer
Na2CO3+H2SO4=Na2SO4+CO2+H2O
m(H2SO4)=150*5/100=7,5г
M(H2SO4)=98г/моль
n(H2SO4)=7,5/98=0,076=0,08моль
nH2SO4 : nCO2=1:1
n(CO2)=0,08моль
V(CO2)=0,08*22,4=1,792л
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Answers & Comments
Verified answer
Na2CO3+H2SO4=Na2SO4+CO2+H2Om(H2SO4)=150*5/100=7,5г
M(H2SO4)=98г/моль
n(H2SO4)=7,5/98=0,076=0,08моль
nH2SO4 : nCO2=1:1
n(CO2)=0,08моль
V(CO2)=0,08*22,4=1,792л