Chemistry - Stoichiometry Question with Solution | TestHub

Solution:

The reaction is: $\mathrm{NaCl}(aq)+{\mathrm{AgNO}}_3(aq)\to \mathrm{AgCl}(s)+{\mathrm{NaNO}}_3(aq)$

Molar mass of NaCl = $22.99+35.45=58.44g/mol$

Molar mass of AgCl = $107.87+35.45=143.32g/mol$

From the stoichiometry, 1 mole of NaCl produces 1 mole of AgCl.

Moles of AgCl precipitated = $\frac{massofAgCl}{molarmassofAgCl}=\frac{10g}{143.32g/mol}=0.06977mol$

Since the mole ratio is 1:1, moles of pure NaCl = $0.06977mol$

Mass of pure NaCl = moles of pure NaCl $\times$ molar mass of NaCl

Mass of pure NaCl = $0.06977mol\times 58.44g/mol=4.076g$

Percentage purity of NaCl = $\frac{massofpureNaCl}{totalmassofimpureNaCl}\times 100$

Percentage purity of NaCl = $\frac{4.076g}{7g}\times 100=58.23\%$