Discussion
Ten grams of calcium carbonate which is only 90% pure is treated with excess hydrochloric acid. What is the mass of CO2 gas liberated? (Atomic mass: Ca=40, C=12 & O=16)
Answer: B
The mass of CO2 is 3.96 g.
Only 90% of the 10 g is CaCO3 (molar mass 100 g mol−1), and each mole gives one mole of CO2 (44 g mol−1):
\[ \begin{aligned} \mathrm{CaCO_{3} + 2HCl} &\rightarrow \mathrm{CaCl_{2} + H_{2}O + CO_{2}} \\ n &= \frac{9}{100} = 0.09\ \text{mol} \\ m(\mathrm{CO_{2}}) &= 0.09 \times 44 = 3.96\ \text{g} \end{aligned} \]
4.4 g ignores the impurity, and 0.44 g and 0.22 g misplace the decimal.
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