JEE Main 2020 · Chemistry · Mole Concept and Stoichiometry · medium

A 10.0 g mixture of CaCO3 and MgCO3 is heated strongly until no more gas is evolved and the solid residue (CaO+MgO) weighs 5.27 g. Taking molar…

Single correct option

10.0\ \mathrm{g}

mixture of

\mathrm{CaCO_3}

and

\mathrm{MgCO_3}

is heated strongly until no more gas is evolved and the solid residue (

\mathrm{CaO}+\mathrm{MgO}

) weighs

5.27\ \mathrm{g}

. Taking molar masses (g mol

^{-1}

)

\mathrm{CaCO_3}=100

\mathrm{MgCO_3}=84

\mathrm{CaO}=56

\mathrm{MgO}=40

, the mass percentage of

\mathrm{CaCO_3}

in the mixture is closest to

Show answer and solution

Answer: (C) $60\%$

Let

x

g be

\mathrm{CaCO_3}

. Residue

=\dfrac{56x}{100}+\dfrac{40(10-x)}{84}=0.56x+4.762-0.4762x=4.762+0.0838x

. Setting this equal to

5.27

gives

x=\dfrac{0.508}{0.0838}\approx 6.0

g, so

\mathrm{CaCO_3}\approx 60\%

. Check:

3.36+1.905=5.265

g. Shortcut: a 50:50 mixture would leave

5.18

g, so the answer must be above 50%.

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