Question

JEE MAINS PRACTISE QUESTION
A 2g mixture of NaHCO3 and Na2CO3 is heated at high temperature abserved loss in weight is 0.25g. Calculate the percentage of Na2CO3 in initial mixture.
Answer properly.
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Answer 1

Answer:

Step-by-step explanation:

Na  

2

​

CO  

3

​

 does not decompose on heating.

Two moles of  NaHCO  

3

​

 decomposes on heating to give two moles of CO  

2

​

.

11.2 L of  CO  

2

​

 corresponds to 0.5 moles of NaHCO  

3

​

 (molecular weight is 84 g/mol) which corresponds to 42 g.

Hence, the percentage of Na  

2

​

CO  

3

​

 in the mixture is  

100

100−42

​

×100=58%.