Question

3.75g of a mixture of $CaCO_3$ and $MgCO_3$ is dissolved in 1L, 0.1N $HCl$ to liberate 0.04 moles of $CO_2$. Calculate The Percentage composition of $CaCO_3$ and $MgCO_3$ in the mixture,

Answer 1

CaCO₃: 65%, MgCO₃: 35%

Answer 2

1. Reactions: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$ $MgCO_3 + 2HCl \rightarrow MgCl_2 + H_2O + CO_2$

2. Mass and Mole Equations: Let $w_{CaCO_3}$ and $w_{MgCO_3}$ be masses. Molar masses: $M_{CaCO_3} \approx 100$ g/mol, $M_{MgCO_3} \approx 84$ g/mol. $w_{CaCO_3} + w_{MgCO_3} = 3.75$ g. $\frac{w_{CaCO_3}}{100} + \frac{w_{MgCO_3}}{84} = 0.04$ moles.

3. Solve Equations: Substitute $w_{MgCO_3} = 3.75 - w_{CaCO_3}$ into the second equation: $\frac{w_{CaCO_3}}{100} + \frac{3.75 - w_{CaCO_3}}{84} = 0.04$ $84 w_{CaCO_3} + 100(3.75 - w_{CaCO_3}) = 0.04 \times 8400$ $84 w_{CaCO_3} + 375 - 100 w_{CaCO_3} = 336$ $-16 w_{CaCO_3} = -39$ $w_{CaCO_3} = 2.4375$ g. $w_{MgCO_3} = 3.75 - 2.4375 = 1.3125$ g.

4. Percentage Composition: % $CaCO_3 = \frac{2.4375}{3.75} \times 100 = 65\%$. % $MgCO_3 = \frac{1.3125}{3.75} \times 100 = 35\%$.