Question

When sample of baking soda is strongly ignited in crucible it suffered loss in weight of 3.1g the mass of baking soda is-
A: 16.8g
B: 8.4g
C: 11.6g
D: 4.2g

Answer 1

Hint: Baking soda also known as sodium hydrogen carbonate or sodium bicarbonate possess the chemical formula,\[NaHC{O_3}\]. This salt comprises of a sodium cation (Na+) and a bicarbonate anion (\[HC{O_3}^ - \]).\[NaHC{O_3}\]is a white solid which is crystalline, but generally appears as in a fine powder form.

Complete step by step answer:
It is given that when a sample of baking soda is strongly ignited in crucible, it suffers loss in weight. Actually, when baking soda is heated above temperature of 80°C, it begins to decompose, generating sodium carbonate (\[N{a_2}C{O_3}\]), water (\[{H_2}O\]) and carbon dioxide (\[C{O_2}\]). The balanced chemical reaction for the ignition of baking soda (\[NaHC{O_3}\]) can be written in the following manner:
\[2NaHC{O_3} \to N{a_2}C{O_3} + {H_2}O + C{O_2}\]
The above balanced chemical equation depicts that loss in weight of baking soda (\[NaHC{O_3}\]) can be due to the generation of water (\[{H_2}O\]) as well as carbon dioxide (\[C{O_2}\]).
2 moles of baking soda or sodium bicarbonate (\[NaHC{O_3}\]) produces 1 mole of water (\[{H_2}O\]) and 1 mole of carbon dioxide (\[C{O_2}\]).
$1 \times 18 = 18$\[\begin{array}{*{20}{l}}
  {Molecular{\text{ }}wt{\text{ }}of{\text{ }}NaHC{O_3} = {\text{ }}84g} \\
  {2{\text{ }}moles{\text{ }}of{\text{ }}NaHC{O_3} = {\text{ }}2 \times 84{\text{ }} = {\text{ }}168g} \\
  {Similarly,} \\
  {Molecular{\text{ }}wt.{\text{ }}of{\text{ }}{H_2}O{\text{ }} = {\text{ }}18g} \\
  {1{\text{ }}mole{\text{ }}of{\text{ }}{H_2}O{\text{ }} = {\text{ }}18g} \\
  {Molecular{\text{ }}wt.{\text{ }}of{\text{ }}C{O_2} = {\text{ }}44g} \\
  {1{\text{ }}mole{\text{ }}of{\text{ }}C{O_2} = {\text{ }}44g}
\end{array}\]
For simplicity, refer to the table below:

Name of molecule	Number of moles involved	Molecular weight
\[NaHC{O_3}\]	2	$2 \times 84 = 168$
\[{H_2}O\]	1	$1 \times 18 = 18$
\[C{O_2}\]	1	$1 \times 44 = 44$

Thus, ignition of \[168\;g{\text{ }}of{\text{ }}NaHC{O_3}\] results in the loss in weight of:
\[{H_2}O{\text{ }} + {\text{ }}C{O_2} = {\text{ }}18{\text{ }} + {\text{ }}44{\text{ }} = {\text{ }}62\;g\].
Hence, loss in weight of \[3.1{\text{ }}g\]of \[NaHC{O_3}\] will correspond to $\dfrac{{168}}{{62}} \times 3.1 = 8.4g$ of \[NaHC{O_3}\].
Thus, the correct answer is Option B i.e. 8.4g

Note: The given baking soda ignition reaction is generally employed in cooking in which \[C{O_2}\] gas leads to several products to rise. As the temperature of the mixture rises, the reaction becomes faster.