Question

An impure sample of solid Na₂CO₃ is allowed to react with 0.1755 M HCl.

Answer 1

Answer: Thus percentage purity of sodium carbonate is 60.83%.

Explanation:

$Na_2CO_3 + 2HCl \longrightarrow 2NaCl + CO_2 + H_2O$

Volume of the HCl solution ,V = 15.55 mL= 0.01555 L

Concentration of HCl solution ,C= 0.1755 M

Moles of HCl in 0.1755 M solution = n

$Molarity=\frac{n}{V(L)}$

$n=C\time V=0.1755 M\times 0.01555 L = 2.729\times 10^{-3} mol$

According o reaction 2 moles of HCl reacts with 1 mole of sodium carbonate.

Then n moles of HCl will react with:

$\frac{1}{2}\times n=\frac{1}{2}\times 2.729\times 10^{-3} mol=1.3645\times 10^{-3} mol$ of sodium carbonate.

Mass of $1.364\times 10^{-3} mol$ of sodium carbonate:

$1.364\times 10^{-3} mol\times 106 g/mol=0.144584 g$

Mass of the sample given= 0.2377 grams

Mass used up in reaction = 0.144585 g

percentage purity= $\frac{0.144585}{0.2337}\times 100=60.83\%$

Thus percentage purity of sodium carbonate is 60.83%.