Mole ConceptMedium
Question
A quantity of 10 g of a piece of marble was put into excess of dilute HCl acid. When the reaction was complete, 1120 cm3 of CO2 was obtained at 0o C and 1 atm. The percentage of CaCO3 in the marble is
Options
A.5%
B.25%
C.50%
D.2.5%
Solution
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
100 g 22400 cm3 at 0oC and 1 atm
$\frac{100}{22400} \times 1120 = 5g \leftarrow 1120\text{ c}\text{m}^{3}$
Hence, percentage of CaCO3 in the marble = $\frac{5}{10} \times 100 = 50\%$
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