Question

The identity of an unknown alkali metal carbonate needs to be obtained. $1.379g$ of the unknown is reacted with $100mL$ of $0.300M$ calcium chloride. $0.986g$ of calcium carbonate was obtained.
Which of the following alkali metal carbonates could be the unknown?

• A. Lithium carbonate.
• B. Potassium carbonate.
• C. Sodium carbonate.
• D. Cesium carbonate.

Answer 1

Answer: (B)

Potassium carbonate.

$M_{2}C{O}_3+CaC{l}_2⟶CaC{O}_3+2MCl$

as $1$ mole of $M_{2}C{O}_3$ gives $1$ mole of $CuC{O}_3$

Moles of $CuC{O}_3$ formed= $\frac{0.986}{100}=9.86\times {10}^{-3}$

Thus of moles of $M_{2}C{O}_3$ reacts= $9.86\times {10}^{-3}$

Mass of $M_{2}C{O}_3$ react= $1.379g$

$\therefore 9.86\times {10}^{-3}$ moles= $\frac{1.379g}{2\times x+60}$

On solving we get

$\Rightarrow x=39.96\simeq 40$ which is molar mass of Potassium.

Thus the unknown carbonate is Potassium Carbonate.