Question

A sample of $NaClO_3$ is converted by heat to NaCl with a loss of $0.16\, g$ of oxygen. The residue is dissolved in water and precipitated as $AgCl$. The mass of $AgCl$ (in g) obtained will be : (Given : Molar mass of $AgCl = 143.5\, g\, mol^{-1}$)

• A. 0.35
• B. 0.41
• C. 0.48
• D. 0.54

Answer 1

Answer: (C)

0.48

Answer 2

The reaction is $2 NaClO _{3} \stackrel{\Delta}{\longrightarrow} 2 NaCl +3 O _{2}$
$3\, mol$ of $O _{2}=2\, mol$ of $NaCl =2\, mol$ of $AgCl$

Number of moles of $O _{2}=\frac{0.16}{32}=0.005 mol$
$0.005\, mol$ of $O _{2}=0.005 \times \frac{2}{3}=0.0033\, mol$ of $AgCl$

The mass of $AgCl$ obtained will $be =143.5 \times 0.0033=0.48\, g$