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 Reactions:

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

Step 1: Define initial masses

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

The total initial mass is \( x + y = 2.21 \, \text{gm} \quad \text{(1)} \)

Step 2: Calculate product yields based on stoichiometry

For \( \text{CaCO}_3 \): 100 gm yields 56 gm of \( \text{CaO} \). The mass of \( \text{CaO} \) produced from \( x \) gm of \( \text{CaCO}_3 \) is \( \frac{56}{100} \times x \).

For \( \text{MgCO}_3 \): 84 gm yields 40 gm of \( \text{MgO} \). The mass of \( \text{MgO} \) produced from \( y \) gm of \( \text{MgCO}_3 \) is \( \frac{40}{84} \times y \).

Step 3: Determine the total weight of the residue

The weight of the residue is the sum of the masses of \( \text{CaO} \) and \( \text{MgO} \) produced. Weight of residue = \( \frac{56 \times x}{100} + \frac{40 \times y}{84} = 1.152 \, \text{gm} \quad \text{(2)} \)

Step 4: Solve the system of equations

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

Step 5: Calculate the total moles of CO₂ formed

The moles of \( \text{CO}_2 \) formed is equal to the sum of the moles of \( \text{CaCO}_3 \) and \( \text{MgCO}_3 \) that reacted.

Moles of \( \text{CO}_2 \) = \( \frac{\text{mass of CaCO}_3}{\text{molar mass of CaCO}_3} + \frac{\text{mass of MgCO}_3}{\text{molar mass of MgCO}_3} \) \[ = \frac{1.19}{100} + \frac{1.02}{84} = 0.0119 + 0.01214 = 0.02404 \, \text{moles} \]

Step 6: Calculate the volume of CO₂ at STP

The molar volume of a gas at STP is 22.4 L/mol or 22400 mL/mol. Volume of \( \text{CO}_2 \) at STP = Moles of \( \text{CO}_2 \) \( \times \) Molar volume at STP \[ V = 0.02404 \, \text{moles} \times 22400 \, \text{mL/mol} = 538.496 \, \text{mL} \]

Final Answer:

\[ \boxed{538.5 \, \text{mL}} \]