Question

Thermal decomposition of \(2.51\,g\) of an impure sample of \(ZnCO_3\) produces \(0.018\,mol\) of \(CO_2\). The percentage of impurity in \(ZnCO_3\) sample is [Atomic mass of Zn = 65.5]

• A. \(1\%\)
• B. \(10\%\)
• C. \(50\%\)
• D. \(90\%\)

Hint:

For decomposition reactions, first use the balanced equation to find moles of reactant decomposed. Then calculate the mass of pure substance and compare it with the given sample mass to determine impurity percentage.

Answer 1

The Correct Option is B

Concept: On heating, zinc carbonate decomposes as \[\begin{aligned} ZnCO_3 \rightarrow ZnO + CO_2 \end{aligned}\] From the balanced equation, \[\begin{aligned} 1\;mol\;ZnCO_3 \rightarrow 1\;mol\;CO_2 \end{aligned}\] Hence, moles of \(ZnCO_3\) decomposed are equal to the moles of \(CO_2\) produced.

Step 1: Calculate the molar mass of \(ZnCO_3\). \[\begin{aligned} M(ZnCO_3) &=65.5+12+3(16)\\ &=125.5\,g\,mol^{-1} \end{aligned}\]

Step 2: Calculate the mass of pure \(ZnCO_3\). Given, \[\begin{aligned} n(CO_2)=0.018\,mol \end{aligned}\] Therefore, \[\begin{aligned} n(ZnCO_3)=0.018\,mol \end{aligned}\] Mass of pure \(ZnCO_3\) \[\begin{aligned} m &=0.018\times125.5\\ &=2.259\,g \end{aligned}\]

Step 3: Calculate the mass of impurity. \[\begin{aligned} \text{Impurity} &=2.51-2.259\\ &=0.251\,g \end{aligned}\]

Step 4: Calculate percentage impurity. \[\begin{aligned} \%\text{ Impurity} &=\frac{0.251}{2.51}\times100\\ &=10\% \end{aligned}\] \[\begin{aligned} \boxed{10\%} \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.