JEE Main 2021 · Chemistry · Mole Concept and Stoichiometry · medium

A 9.2 g mixture of CaCO3 and MgCO3 is strongly heated until only the oxides remain; the residue weighs 4.8 g. The mass percentage of CaCO3 in the…

Single correct option

9.2\ \mathrm{g}

mixture of

\mathrm{CaCO_3}

and

\mathrm{MgCO_3}

is strongly heated until only the oxides remain; the residue weighs

4.8\ \mathrm{g}

. The mass percentage of

\mathrm{CaCO_3}

in the mixture is (Ca = 40, Mg = 24, C = 12, O = 16)

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Answer: (B) $54.3\%$

Mass lost is

\mathrm{CO_2}

9.2-4.8=4.4\ \mathrm{g}=0.1

mol. Each mole of carbonate releases one mole of

\mathrm{CO_2}

, so

x+y=0.1

where

x,y

are moles of

\mathrm{CaCO_3}

and

\mathrm{MgCO_3}

. Mass of mixture:

100x+84y=9.2

. Substituting

y=0.1-x

16x+8.4=9.2

, so

x=0.05

mol. Mass of

\mathrm{CaCO_3}=5.0\ \mathrm{g}

, which is

5.0/9.2\times100=54.3\%

. Check: oxides

=0.05(56)+0.05(40)=4.8\ \mathrm{g}

. The

45.7\%

option is the

\mathrm{MgCO_3}

share.

Original question written in the JEE Main 2021 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake