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

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.