Question

A sample of \( \text{CaCO}_3 \) and \( \text{MgCO}_3 \) weighed 2.21 g is ignited to constant weight of 1.152 g. The composition of the mixture is:
(Given molar mass in g mol\(^{-1}\))

• A. \( 1.187 \, \text{g} \, \text{CaCO}_3 + 1.023 \, \text{g} \, \text{MgCO}_3 \)
• B. \( 1.023 \, \text{g} \, \text{CaCO}_3 + 1.023 \, \text{g} \, \text{MgCO}_3 \)
• C. \( 1.187 \, \text{g} \, \text{CaCO}_3 + 1.187 \, \text{g} \, \text{MgCO}_3 \)
• D. \( 1.023 \, \text{g} \, \text{CaCO}_3 + 1.187 \, \text{g} \, \text{MgCO}_3 \)

Answer 1

The Correct Option is A

Given Reaction:

\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \] \[ \text{MgCO}_3 \rightarrow \text{MgO} + \text{CO}_2 \]

Step 1: Relating masses of CaCO₃ and CaO

Let the mass of \( \text{CaCO}_3 \) be \( x \, \text{gm} \) and the mass of \( \text{MgCO}_3 \) be \( y \, \text{gm} \).

\[ \Rightarrow (x + y) = 2.21 \, \text{gm} \quad \text{(1)} \]

Step 2: Proportions based on given reactions

100 gm of \( \text{CaCO}_3 \) gives 56 gm of \( \text{CaO} \), so: \[ \frac{x}{100} = \frac{56}{100} \quad \Rightarrow \quad x = \frac{56}{100} \times x \]

Similarly, 84 gm of \( \text{MgCO}_3 \) gives 40 gm of \( \text{MgO} \): \[ y = \frac{40}{84} \times y \]

Step 3: Calculating the weight of the residue

\[ \text{Wt. of residue} = \frac{56 \times x}{100} + 40 \times y \quad \Rightarrow \quad 1.152 \]

Step 4: Solving equations (1) and (2)

Solving equations (1) and (2), we get: \[ x = 1.19 \quad \text{gm}, \quad y = 1.02 \, \text{gm} \]

Step 5: Mole calculation

Mole of \( \text{CO}_2 \) formed = Mole of \( \text{CaCO}_3 \) + Mole of \( \text{MgCO}_3 \)

\[ = \frac{1.19}{100} + \frac{1.02}{84} = 0.0241 \, \text{moles} \]

Step 6: Volume of \( \text{CO}_2 \) at STP

Volume of \( \text{CO}_2 \) at STP: \[ V = 0.0421 \times 22.4 = 539.8 \, \text{mL} \]

Final Answer:

\[ \boxed{539.8 \, \text{mL}} \]

Answer 2

Reactions:

\[ \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \] \[ \text{MgCO}_3(s) \rightarrow \text{MgO}(s) + \text{CO}_2(g) \]

Let the weight of \(\text{CaCO}_3\) be \(x\) g. Then, the weight of \(\text{MgCO}_3\) is \((2.21 - x)\) g.

Moles of \(\text{CaCO}_3\) decomposed = moles of \(\text{CaO}\) formed

\[ \frac{x}{100} = \text{moles of \(\text{CaO}\) formed.} \]

Weight of \(\text{CaO}\) formed:

\[ \text{weight of \(\text{CaO}\) formed} = \frac{x}{100} \times 56. \]

Moles of \(\text{MgCO}_3\) decomposed = moles of \(\text{MgO}\) formed

\[ \frac{2.21 - x}{84} = \text{moles of \(\text{MgO}\) formed.} \]

Weight of \(\text{MgO}\) formed:

\[ \text{weight of \(\text{MgO}\) formed} = \frac{2.21 - x}{84} \times 40. \]

According to the problem:

\[ \frac{2.21 - x}{84} \times 40 + \frac{x}{100} \times 56 = 1.152. \]

Solving, we find:

\[ x = 1.1886 \, \text{g (weight of \(\text{CaCO}_3\))}. \]

And the weight of \(\text{MgCO}_3\) is:

\[ 2.21 - x = 1.0214 \, \text{g}. \]