\( 1.84 gm \) mixture of \( CaCO _{3} \) & \( MgCO 3 \) exactly produc

1.	\( 1.84 gm \) mixture of \( CaCO _{3} \) & \( MgCO 3 \) exactly produces 0.88 g CO2 on strongly heating, calculate \( \% \) composition of original mixture.
Answer» Let say x.g CaCO₃ is present in mixture. ∴ Weight of MgCO₃ in mixture = (1.84 - x)g. \(\underset{1 mol}{CaCO_3} \overset{\Delta}\longrightarrow \underset{1 mol}{CaO} + \underset{1 mol} {CO_2}\) ∴ Number of moles of CO₂ produced by CaCO₃ = \(\frac x{100}\) mol \(\underset{1 mol}{MgCO_3} \overset{\Delta}\longrightarrow \underset{1 mol}{MgO} + \underset{1 mol}{CO_2}\) ∴ Number of mole of CO₂ produced by MgCO₃= \(\frac{1.84 - x}{84}\) mol But we have given 0.88g CO₂ produced. ∴ \(\left(\frac x{100} + \frac{1.84 - x}{84}\right)\times 44 = 0.88 g\) ∴ \(\frac x{100} + \frac{1.84 - x}{84} = 0.02\) \(84x + 184 - 100 x = 0.02 \times 8400\) \(184 - 16x = 168\) \(16x = 184 - 168\) \(16x = 16\) \(x = 1 g\) ∴ Weight of CaCO₃ in mixture = 1g and Weight of MgCO₃ = (1.84 - 1)g ∴ % composition of CaCO₃ = \(\frac1{1.84} \times 100\) = \(54.35\)% ∴ % composition of MgCO₃ = \(\frac{0.84}{1.84} \times 100\) = \(45.65\)%

\( 1.84 gm \) mixture of \( CaCO _{3} \) & \( MgCO 3 \) exactly produces 0.88 g CO2 on strongly heating, calculate \( \% \) composition of original mixture.

Answer»

Let say x.g CaCO₃ is present in mixture.

∴ Weight of MgCO₃ in mixture = (1.84 - x)g.

\(\underset{1 mol}{CaCO_3} \overset{\Delta}\longrightarrow \underset{1 mol}{CaO} + \underset{1 mol} {CO_2}\)

∴ Number of moles of CO₂ produced by CaCO₃ = \(\frac x{100}\) mol

\(\underset{1 mol}{MgCO_3} \overset{\Delta}\longrightarrow \underset{1 mol}{MgO} + \underset{1 mol}{CO_2}\)

∴ Number of mole of CO₂ produced by MgCO₃= \(\frac{1.84 - x}{84}\) mol

But we have given 0.88g CO₂ produced.

∴ \(\left(\frac x{100} + \frac{1.84 - x}{84}\right)\times 44 = 0.88 g\)

∴ \(\frac x{100} + \frac{1.84 - x}{84} = 0.02\)

\(84x + 184 - 100 x = 0.02 \times 8400\)

\(184 - 16x = 168\)

\(16x = 184 - 168\)

\(16x = 16\)

\(x = 1 g\)

∴ Weight of CaCO₃ in mixture = 1g

and Weight of MgCO₃ = (1.84 - 1)g

∴ % composition of CaCO₃ = \(\frac1{1.84} \times 100\) = \(54.35\)%

∴ % composition of MgCO₃ = \(\frac{0.84}{1.84} \times 100\) = \(45.65\)%