Question

$70\mathrm{g}$ of a mixture of sodium carbonate and sodium bicarbonate is taken and subjected to heating until there is no further loss in weight. The heating was followed by condensation. The loss in weight of the sample was $15.7\%$ of the initial weight of the mixture. Find out the amount of the sodium carbonate in the mixture.

• A. $42\mathrm{g}$
• B. $53\mathrm{g}$
• C. $28\mathrm{g}$
• D. $64\mathrm{g}$

Answer 1

The correct option is A

$42\mathrm{g}$

• There are two compounds in the mixture i.e., Sodium carbonate (${\mathrm{Na}}_2{\mathrm{CO}}_3$) and Sodium bicarbonate (${\mathrm{NaHCO}}_3$).
• The total amount of mixture is $70\mathrm{g}$.

Explanation for the correct answer:

Let the mass of Sodium carbonate (${\mathrm{Na}}_2{\mathrm{CO}}_3$) = $\mathrm{x}\mathrm{gm}$

And the mass of Sodium bicarbonate (${\mathrm{NaHCO}}_3$) will be = $(70-x)\mathrm{gm}$

Loss in weight on heating will be due to the decomposition of ${\mathrm{NaHCO}}_3$

After decomposition, the weight of the remaining substance =

$70\mathrm{gm}-\frac{15.7}{100}\times 70\mathrm{gm}=59\mathrm{gm}$

The chemical reactions involved will be:

${\mathrm{Na}}_2{\mathrm{CO}}_3\stackrel{\Delta }{⟶}\text{no effect}$

$\underset{\frac{2-x}{84}\text{mole}}{2{\mathrm{NaHCO}}_3}\stackrel{\Delta }{⟶}\underset{\frac{2-x}{2\times 84}\text{mole}}{{\mathrm{Na}}_2{\mathrm{CO}}_3}+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

As we know the molecular mass of ${\mathrm{Na}}_2{\mathrm{CO}}_3=106$ gm

$x+\left(\frac{(2-x)}{84\times 2}\times 106\right)=59$

$84\mathrm{x}+3710-53\mathrm{x}=4956$

$\mathrm{x}=\frac{4956-3710}{31}=41.7~42\mathrm{gm}$

So, the amount of Sodium carbonate in the mixture is $42\mathrm{gm}$.

Therefore, the correct answer is an option (A).