Question

the weight of caco3 is required to produce carbondioxide that is sufficient for conversion of 1 0.1 mole sodium carbonate to sodium bicarbonate is ​

Answer 1

Answer:

Explanation: molecular mass of Na2CO3=106gm, that of CO2=44gm and of CaCO3=100gm If x gm CO2 react completely with sod carbonate --> x=(21.2X44)/106= 8.8 kg and this 8.8kg CO2 evolve only when 20 kg CaCO3 is decomposed (using rule of proportion or unitary method).

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Answer 2

Answer:

20 kg

CaCO3→CaO+CO2.

100−56→44 

Na2CO3+CO2+H2O→2NaHCO3

106+18+44→2×84

Now you have 21.2 kg so 21200×10644.

= 8.8 kg of CO2 needed.

From eq.1  88000×44100.

= 20 kg of CaCO3 is needed.