JEE Main 2023 · Chemistry · Mole Concept and Stoichiometry · medium

A 2.00 g mixture of CaCO3 and MgCO3 is strongly heated until both decompose completely to oxides, releasing 0.947 g of CO2. Taking molar masses…

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2.00\ \mathrm{g}

mixture of

\mathrm{CaCO_3}

and

\mathrm{MgCO_3}

is strongly heated until both decompose completely to oxides, releasing

0.947\ \mathrm{g}

\mathrm{CO_2}

. Taking molar masses

\mathrm{CaCO_3}=100

\mathrm{MgCO_3}=84

and

\mathrm{CO_2}=44\ \mathrm{g\,mol^{-1}}

, the mass percentage of

\mathrm{CaCO_3}

in the mixture is closest to

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Answer: (C) $60\%$

Let

\mathrm{CaCO_3}=x

g, so

\mathrm{MgCO_3}=2-x

g. Each mole of carbonate gives one mole of

\mathrm{CO_2}

, so

44\left(\frac{x}{100}+\frac{2-x}{84}\right)=0.947

. Hence

\frac{x}{100}+\frac{2-x}{84}=0.02152

. Multiplying by 8400:

84x+100(2-x)=180.8

, i.e.

200-16x=180.8

and

x=1.2

g. Mass percentage

=1.2/2.00\times100=60\%

. Shortcut: if the sample were pure

\mathrm{MgCO_3}

it would give

1.048

g and if pure

\mathrm{CaCO_3}

0.88

g, so the observed value lies about

60\%

of the way towards the lighter-yielding

\mathrm{CaCO_3}

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