JEE Main 2025 · Chemistry · Mole Concept and Stoichiometry · medium

A 1.84 g mixture of CaCO3 and MgCO3 (molar masses 100 and 84 g mol^-1) on strong heating liberates 0.448 L of CO2 measured at STP (22.4 L mol^-1).…

Single correct option

1.84\ \mathrm{g}

mixture of

\mathrm{CaCO_3}

and

\mathrm{MgCO_3}

(molar masses

100

and

84\ \mathrm{g\,mol^{-1}}

) on strong heating liberates

0.448\ \mathrm{L}

\mathrm{CO_2}

measured at STP (

22.4\ \mathrm{L\,mol^{-1}}

). The mass percentage of

\mathrm{CaCO_3}

in the mixture is

Show answer and solution

Answer: (C) $54.3\%$

n(\mathrm{CO_2}) = 0.448/22.4 = 0.02

mol, and each mole of carbonate gives one mole of

\mathrm{CO_2}

, so

x+y=0.02

where

x,y

are the moles of

\mathrm{CaCO_3}

\mathrm{MgCO_3}

. Mass:

100x+84y=1.84

. Substituting

y=0.02-x

16x = 1.84-1.68 = 0.16

, so

x=0.01

mol. Mass of

\mathrm{CaCO_3}=1.00

g, giving

1.00/1.84 \times 100 = 54.3\%

. Option

45.7\%

is the

\mathrm{MgCO_3}

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