Calculate the mass of \(\mathrm{ZnSO}_4\). The equation for the reaction is ____________ \(\mathrm{ZnCO}_3+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{ZnSO}_4+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\) \((\mathrm{Zn}=65, \mathrm{~S}=32, \mathrm{O}=16)\)
A. \(32.2\) B. \(37.5\) C. \(38.6\) D. \(34.3\)
Correct Answer: A
Explanation
\(\mathrm{ZnCO}_3+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{ZnSO}_4+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2\) mole of \(\mathrm{H}_2 \mathrm{SO}_4=4\) moldm \(^{-3} \times\left(\frac{50}{1000}\right) \mathrm{dm}^3\) \(=0.2 \mathrm{~mol}\).rom the balanced equation of reaction, 1 mole of \(\mathrm{H}_2 \mathrm{SO}_4\) produces 1 mole of \(\mathrm{ZnSO}_4\). Thus, \(0.2 \mathrm{~mol}\) of \(\mathrm{H}_2 \mathrm{SO}_4\) will generate \(0.2\) mole of \(\mathrm{ZnSO}_4\). molar mass of \(\mathrm{ZnSO}_4\) \(=[65+32+(16 \times 4)] \mathrm{gmol}^{-1}=161 \mathrm{gmol}^{-1}\). Mass of \(\mathrm{ZnSO}_4=(0.2 \times 161) \mathrm{g}=32.2 \mathrm{~g}\)