wat is the composition of mixture containing KClO3,K2CO3,KHCO3,KCL was heated producing CO2,O2,H2O?

Question

Equations -

2KClO3 ---> 2KCl +3O2
2KHCO3 --->K2O+H2O+2CO2
K2CO3 -----> K2O+ CO2

the KCl does not react under the conditions of the reaction.

If 100g of the mixture produces 1.80g of H2O, 13.20 g of CO2 and 4g of O2,

WAT WAS THE COMPOSITION OF THE ORIGINAL MIXTURE?(assume complete decomposition of the mixture)

P.S.----- please show me the full step..... homework help.... please...

Answer 1

Molecular weights: CO2=44 O2=32 H2O=18 KClO3=122.5 K2CO3=138 KHCO3=100

KHCO3 and K2CO3 both give CO2, so leave CO2 for awhile. Start with H2O:

1.8gH2O x 1molH2O/18gH2O x 2molKHCO3/1molH2O x 100gKHCO3/1molKHCO3 = 20g KHCO3 in the original 100g mixture.

4gO2 x 32gO2/1molO2 x 2molKClO3/3molO2 x 122.5KClO3/1molKClO3 = 10g KClO3 in the 100g mixture

Solve for the amount of CO2 from the 20g KHCO3 only:

20gKHCO3 x 1molKHCO3/100KHCO3 x 2molCO2/2molKHCO3 = 0.2 moles CO2

Find total moles produced:

13.20gCO2 x 1molCO2/44gCO2 = 0.3mol CO2

So 0.1 mole CO2 comes from K2CO3:

0.1molCO2 x 1molK2CO3/1molCO2 x 138gK2CO3/1molK2CO3 = 14g K2CO3 in the 100g mixture.

20gKHCO3 + 10gKClO3 + 14gK2CO3 = 44g

So 100 - 44 = 56g KCl