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

Step 1: Write the decomposition.
On heating, zinc carbonate breaks down as
\[ ZnCO_3 \rightarrow ZnO + CO_2 \]
So one mole of $ZnCO_3$ gives one mole of $CO_2$.

Step 2: Find moles of pure $ZnCO_3$.
Since the mole ratio is $1:1$, the moles of pure $ZnCO_3$ equal the moles of $CO_2$, which is $0.018$ mol.

Step 3: Find the molar mass of $ZnCO_3$.
\[ M = 65.5 + 12 + 3\times16 = 125.5 \]

Step 4: Find the mass of pure $ZnCO_3$.
\[ \text{mass} = 0.018 \times 125.5 = 2.259\ g \approx 2.26\ g \]

Step 5: Find the impurity mass.
The total sample was $2.51\,g$, so impurity $= 2.51 - 2.26 = 0.25\,g$ (about).

Step 6: Convert to percentage.
\[ \%\ \text{impurity} = \frac{0.25}{2.51}\times100 \approx 10\% \]
\[ \boxed{10\%} \]