Question

1.84 g of a mixture of CaCO\(_3\) and MgCO\(_3\) is heated till no further weight loss. The weight of the residue is 0.96 g. The % composition of CaCO\(_3\) in the mixture is ................. \[ \text{(round off to two decimal places)} \] \text{[Given: Atomic weight of Ca = 40; Mg = 24; C = 12; O = 16]}

Hint:

To find the percentage composition of a mixture, use the mass relationship between the reactants and products, then solve for the unknown mass.

Answer 1

Let the mass of CaCO\(_3\) in the mixture be \( x \) g and the mass of MgCO\(_3\) be \( (1.84 - x) \) g.
When heated, both \( \text{CaCO}_3 \) and \( \text{MgCO}_3 \) decompose into their oxides:
- \( \text{CaCO}_3 \) decomposes into \( \text{CaO} \) and \( \text{CO}_2 \).
- \( \text{MgCO}_3 \) decomposes into \( \text{MgO} \) and \( \text{CO}_2 \).
The weight loss during heating corresponds to the loss of \( \text{CO}_2 \) from both compounds. The mass of the residue is the sum of the masses of \( \text{CaO} \) and \( \text{MgO} \).
Step 1: Calculate the masses of CaO and MgO.
- The molar mass of \( \text{CaCO}_3 \) is \( 40 + 12 + 48 = 100 \) g/mol. The molar mass of \( \text{CaO} \) is \( 56 \) g/mol.
- The molar mass of \( \text{MgCO}_3 \) is \( 24 + 12 + 48 = 84 \) g/mol. The molar mass of \( \text{MgO} \) is \( 40 \) g/mol.
For \( x \) g of \( \text{CaCO}_3 \), the mass of \( \text{CaO} \) formed is: \[ \text{Mass of CaO} = \frac{56}{100} \times x \] For \( (1.84 - x) \) g of \( \text{MgCO}_3 \), the mass of \( \text{MgO} \) formed is: \[ \text{Mass of MgO} = \frac{40}{84} \times (1.84 - x) \] Step 2: Set up the equation for the total residue.
The total mass of the residue is 0.96 g, so: \[ \frac{56}{100} \times x + \frac{40}{84} \times (1.84 - x) = 0.96 \] Step 3: Solve for \( x \).
Simplifying the equation: \[ 0.56x + 0.476 \times (1.84 - x) = 0.96 \] \[ 0.56x + 0.87664 - 0.476x = 0.96 \] \[ 0.084x = 0.08336 \] \[ x = \frac{0.08336}{0.084} \approx 0.99 \, \text{g} \] Step 4: Calculate the percentage of CaCO\(_3\).
The percentage of CaCO\(_3\) in the mixture is: \[ \frac{0.99}{1.84} \times 100 \approx 53.8% \] Final Answer: \[ \boxed{53.8} \]