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

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:

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