Question

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

Answer 1

Hint: 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,
\[2NaHC{O_3}\xrightarrow{\Delta }N{a_2}C{O_3} + C{O_2} + {H_2}O\]
Here, two moles of sodium bicarbonate are heated and formed, one mole of sodium carbonate, carbon dioxide and water.
\[NaCl\xrightarrow{\Delta }No\,reaction\]
Sodium chloride will not react with heat.
Molecular mass of sodium bicarbonate\[ = 84g/mol\]
Molecular mass of carbon dioxide \[ = 44g/mol\]
2 moles of sodium bicarbonate is equal to \[168g\]. And here, \[168g\] of \[NaHC{O_3}\] produces, \[44g\,of\,C{O_2}\]
Hence, the amount of \[NaHC{O_3}\] to evolved \[0.66g\,of\,C{O_2}\]\[ = \dfrac{{168x0.66}}{{44}}\]
\[ = 2.52g\]
Therefore, percentage of sodium bicarbonate\[ = \dfrac{{2.52x100}}{4}\]
\[ = 63\% \]
And the percentage of sodium chloride\[ = 100 - 63\]
\[ = 37\% \]
Hence, option (A) is correct.
The percentage composition of the original mixture by mass is not equal to, \[42\% \,and\,58\% \].Hence, the option (B) is incorrect.
The percentage composition of the original mixture by mass is not equal to, \[31.5\% \,and\,68.5\% \]Hence, option (C) is incorrect.
The percentage composition of sodium bicarbonate and sodium chloride is not equal to, \[37\% \,and\,63\% \]. 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, \[NaHC{O_3} = 63\% ,NaCl = 37\% \]