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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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3.36 L

4.48 L

1.12 L

2.24 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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