Question

2g of carbonate, bicarbonate and chloride of sodium when heated at STP produces 56ml of CO2(g). The percentage of sodium bicarbonate in the mixture is?

Answer 1

Dear Student,

Total weight of mixture of Na₂CO₃, NaHCO₃ and NaCl given = w = 2 gm

Na₂CO₃ doesnot compose on normal heating and NaCl on heating doesnot produce CO₂ gas. Only NaHCO₃​ will decompose on heating as follows,

NaHCO_{3$\stackrel{∆}{\to }$} Na₂CO₃ $+$ CO₂ $+$ H₂O
1 ml of CO₂ = 0.001977 gm of CO₂ as per density, so
​56 ml of CO₂ = 0.1107 gm of CO₂

Use molar mass of CO₂ to convert mass to moles,

0.1107 gm $\times$(1 mole of CO₂/44 gm) = 0.002516 moles of CO₂

From reaction, we can see that

2 moles of NaHCO₃ yields 1 mole of CO₂ on heating , therefore ,

0.002516 $\times$ (2 moles of NaHCO₃ / 1 mole of CO₂ ) = 0.005032 moles of NaHCO₃

To convert this into gm, use molar mass of ​NaHCO₃

0.005032 $\times$ (84.007/1)= 0.4261 gm

This means, % composition of ​NaHCO₃ in mixture = (0.4261 /2)$\times$100
= 21.30% ​


Regards.