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Question: When \[4g\]of a mixture of \[NaHC{O_3}\] and \[NaCl\] is heated, \[0.66g\,C{O_2}\]​ gas is evolved. ...

When 4g4gof a mixture of NaHCO3NaHC{O_3} and NaClNaCl is heated, 0.66g CO20.66g\,C{O_2}​ gas is evolved. The percentage composition of the original mixture by mass is-
A.NaHCO3=63%,NaCl=37%NaHC{O_3} = 63\% ,NaCl = 37\%
B.NaHCO3=42%,NaCl=58%NaHC{O_3} = 42\% ,NaCl = 58\%
C.NaHCO3=31.5%,NaCl=68.5%NaHC{O_3} = 31.5\% ,NaCl = 68.5\%
D.NaHCO3=37%,NaCl=63%NaHC{O_3} = 37\% ,NaCl = 63\%

Explanation

Solution

By heating sodium bicarbonate, there is a formation of sodium carbonate with the liberation of carbon dioxide and water. But there is no reaction on heating sodium chloride. And the percentage composition can be found by using the molecular formula. And the percentage composition will help for the chemical analysis of the compound.

Complete answer:
The chemical reaction of heating of sodium bicarbonate can be written as,
2NaHCO3→ΔNa2CO3+CO2+H2O2NaHC{O_3}\xrightarrow{\Delta }N{a_2}C{O_3} + C{O_2} + {H_2}O
Here, two moles of sodium bicarbonate are heated and formed, one mole of sodium carbonate, carbon dioxide and water.
NaCl→ΔNo reactionNaCl\xrightarrow{\Delta }No\,reaction
Sodium chloride will not react with heat.
Molecular mass of sodium bicarbonate=84g/mol = 84g/mol
Molecular mass of carbon dioxide =44g/mol = 44g/mol
2 moles of sodium bicarbonate is equal to 168g168g. And here, 168g168g of NaHCO3NaHC{O_3} produces, 44g of CO244g\,of\,C{O_2}
Hence, the amount of NaHCO3NaHC{O_3} to evolved 0.66g\,of\,C{O_2}$$$$ = \dfrac{{168x0.66}}{{44}}
=2.52g= 2.52g
Therefore, percentage of sodium bicarbonate=2.52x1004 = \dfrac{{2.52x100}}{4}
=63%= 63\%
And the percentage of sodium chloride=100−63 = 100 - 63
=37%= 37\%
Hence, option (A) is correct.
The percentage composition of the original mixture by mass is not equal to, 42% and 58%42\% \,and\,58\% .Hence, the option (B) is incorrect.
The percentage composition of the original mixture by mass is not equal to, 31.5% and 68.5%31.5\% \,and\,68.5\% Hence, option (C) is incorrect.
The percentage composition of sodium bicarbonate and sodium chloride is not equal to, 37% and 63%37\% \,and\,63\% . Hence, the option (D) is correct.

Note:
Sodium chloride is not reacting with heat and it does not give any reaction when it is heated. But by heating sodium bicarbonate, there is a formation of sodium carbonate with the liberation of carbon dioxide and water. And the percentage composition of the original mixture by mass will get as, NaHCO3=63%,NaCl=37%NaHC{O_3} = 63\% ,NaCl = 37\%