JEE Main 2018 · Chemistry · Mole Concept and Stoichiometry · medium

A 2.000 g mixture of CaCO3 and MgCO3 is strongly heated until only the oxides remain; the residue weighs 1.036 g. Taking molar masses CaCO3=100…

Single correct option

2.000\ \mathrm{g}

mixture of

\mathrm{CaCO_3}

and

\mathrm{MgCO_3}

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

1.036\ \mathrm{g}

. Taking molar masses

\mathrm{CaCO_3}=100

\mathrm{MgCO_3}=84

\mathrm{CaO}=56

\mathrm{MgO}=40\ \mathrm{g\,mol^{-1}}

, the mass percentage of

\mathrm{CaCO_3}

in the mixture is closest to

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Answer: (C) $50\%$

Let

x

g be

\mathrm{CaCO_3}

. Residue

=\frac{56}{100}x+\frac{40}{84}(2-x)=1.036

. So

0.56x+0.9524-0.4762x=1.036

, giving

0.0838x=0.0836

and

x\approx 1.00

g. Hence

\mathrm{CaCO_3}

1.00/2.000\times100=50\%

. Shortcut: if the sample were pure

\mathrm{MgCO_3}

the residue would be

0.952

g and pure

\mathrm{CaCO_3}

1.12

1.036

lies exactly midway, so the mixture is about 1:1 by mass.

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