Question

1.84 g of a mixture of CaCO\(_3\) and MgCO\(_3\) is strongly heated to get a residue of 0.96 g. The percentage of CaCO\(_3\) in the mixture is

• A. 50.34
• B. 49.66
• C. 54.34
• D. 45.66

Hint:

When dealing with decomposition reactions, use stoichiometric molar mass changes to find mass relationships. Form equations for residue and solve algebraically.

Answer 1

The Correct Option is C

Step 1: Use decomposition reactions. 
\[ \text{CaCO}_3 \xrightarrow{\Delta} \text{CaO} + \text{CO}_2 \quad (\text{Molar mass} = 100 \rightarrow 56) \] \[ \text{MgCO}_3 \xrightarrow{\Delta} \text{MgO} + \text{CO}_2 \quad (\text{Molar mass} = 84 \rightarrow 40) \] Step 2: Let the mass of CaCO\(_3\) in the mixture be \(x\) g. 
Then mass of MgCO\(_3\) = \(1.84 - x\) g 
Step 3: Calculate residue mass from decomposition. 
Mass of residue = CaO from CaCO\(_3\) + MgO from MgCO\(_3\): \[ \text{Residue} = \frac{56}{100}x + \frac{40}{84}(1.84 - x) \] Step 4: Equate to given residue and solve. \[ \frac{56}{100}x + \frac{40}{84}(1.84 - x) = 0.96 \] Step 5: Solve for \(x\). 
\[ 0.56x + 0.877(1.84 - x) = 0.96 \Rightarrow 0.56x + 1.614 - 0.877x = 0.96 \Rightarrow -0.317x = -0.654 \Rightarrow x \approx 2.062 \quad (\text{Not possible due to mass constraint}) \] Correction in step: Re-check using simplified algebra: \[ \frac{56}{100}x + \frac{40}{84}(1.84 - x) = 0.96 \Rightarrow 0.56x + 0.4762(1.84 - x) = 0.96 \Rightarrow 0.56x + 0.8762 - 0.4762x = 0.96 \Rightarrow 0.0838x = 0.0838 \Rightarrow x = 1.00 \, \text{g} \] Step 6: Calculate percentage. 
\[ \frac{1.00}{1.84} \times 100 \approx 54.34\% \]