Question

the weight of the calcium carbonate required to produce carbon dioxide that is sufficient for conversion of one 0.1 mole sodium carbonate to sodium bicarbonate is ​

Answer 1

Answer:

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Explanation:

ANSWER

CaCO  

3

​  

+H  

2

​  

O+Na  

2

​  

CO  

3

​  

⟶CaO+NaHCO  

3

​  

 

Mw of CaCO  

3

​  

=100

Na  

2

​  

CO  

3

​  

=10.6.

  10.6gm=100×  

10.6

10.6

​  

 

=10gmofCaCO  

3

​  

 

Therefore 10.6gm=100×  

10.6

10.6

​  

 

=10gmofCaCO  

3

​