A mixture containing KClO3,K2CO3,KHCO3, and KCl was heated, producing CO2,O2, an

Question

A mixture containing KClO3,K2CO3,KHCO3, and KCl was heated, producing CO2,O2, and H2O gases according to the following equations:

2KClO3(s)?2KCl(s)+3O2(g)2KHCO3(s)?K2O(s)+H2O(g)+2CO2(g)K2CO3(s)?K2O(s)+CO2(g)
The KCl does not react under the conditions of the reaction. 100.0 g of the mixture produces 1.90 g of H2O, 13.64 of CO2, and 4.10 g of O2. (Assume complete decomposition of the mixture.)

1. How many grams of KClO3 were in the original mixture? ** I got 10.5 g

2. How many grams of KHCO3 were in the original mixture? **I got 21.1 g

3. How many grams of K2CO3 were in the original mixture? ** I CAN'T GET THE RIGHT ANSWER ON THIS. I got 42.83 but MasteringChemistry says I'm wrong

4. How many grams of KCl were in the original mixture? ** I have no gotten this answer yet

Explanation / Answer

3)

2KHCO3(s)----------------------> K2O(s)+H2O(g)+2CO2(g)

200 g KHCO3 = 88g CO2

21.1 g KHCO3 = ?

CO2 weight = 88 x 21.1 / 200

                    = 9.284 g

now

total CO2 weight = 13.64 g

9.284 g released in above equation .

remaining CO2 = 13.64 -9.284

                          = 4.356 g

K2CO3(s)--------------------------> K2O(s)+CO2

138 g = 44g CO2

     ?       = 4.356 g CO2

K2CO3 weight = 138 x 4.356 /44

                         = 13.66g

13.66 grams of K2CO3 were in the original mixture

4)

total weight = 100

weight of

KClO3 + KHCO3 + K2CO3 + KCl =100

10.5 + 21.1 + 13.66 + KCl = 100

KCl = 54.7 g

54.7 grams of KCl were in the original mixture

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