NCACO3 = 0.1 MOL MCO2 + MH2O = 16,8 GAM => NCA(HCO3)2 = NCA(...

Câu 23

n

CaCO3

= 0.1 mol

m

CO2

+ m

H2O

= 16,8 gam

=> n

_Ca(HCO3)2

= n

_Ca(OH)2

- n

_CaCO3

= 0, 1 mol

=> n

C

= 2 x n

Ca(HCO3)2

+ n

CaCO3

= 0,3 mol

=> m

CO2

= 13,2 gam

=> m

_H2O

= 3.6 gam

=> n

H2O

= 0,2 mol

n

C

= 0.3 mol, n

H

= 0, 4 mol

=> X la C

3

H

₄