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18.4 gm of a mixture of CaCO₃ and MgCO₃ on heating gives 4 gm of magnesium oxide. The volume of CO₂ produced at STP in this process

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1.12 L

4.48 L

2.24 L

3.36 L

answer is B.

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Detailed Solution

Weight of MgCO₃ and CaCO₃ mixture = 18.4g

Weight of MgO obtained on heating = 4g

$large mathop {mathop {MgC{O_3}(s)}limits_{84g}^{1mole} }limits_{'X'g} xrightarrow{Delta }mathop {mathop {MgO(s)}limits_{40g}^{1mole} }limits_{4g} + C{O_2}(g)$

$large X = frac{4}{40} times 84 = 8.4g$

Weight of MgCO₃ in the mixture = 8.4g

Weight of CaCO₃ in the mixture = (18.4 - 8.4)g = 10g

I $large mathop {MgC{O_3}(s)}limits_{8.4g}^{84g} xrightarrow{}MgO(s) + mathop {C{O_2}(g)}limits_{'{X_1}'l}^{22.4l}$

$large X_1 = frac{8.4}{84} times 22.4 = 2.24l$

II $large mathop {CaC{O_3}(s)}limits_{10g}^{100g} xrightarrow{{}}CaO(s) + mathop {C{O_2}(g)}limits_{'{X_2}'l}^{22.4l}$

$large X_2 = frac{100}{10} times 22.4 = 2.24l$

Total volume of CO₂ liberated at STP = 2.24 + 2.24

= 4 . 48 L

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