Question

3.68 of a mixture of $CaCO _{3}$ and $MgCO _{3}$ is heated to liberate 0.04 mole of $CO _{2}$. The mole $\%$ of $CaCO _{3}$ and $MgCO _{3}$ in the mixture is respectively:

• A. $50 \%, 50 \%$
• B. $60 \%, 40 \%$
• C. $40 \%, 60 \%$
• D. $30 \%, 70 \%$

Answer 1

Answer: (A)

Let the mass of $CaCO _{3}= x\,g$

Then, mass of $MgCO _{2}=(3.68- x ) g$

moles of $CaCO _{3}=\frac{ x }{100}$

moles of $MgCO _{3}=\frac{3.68- x }{84}$

Applying POAC for C-atoms

$\frac{x}{100}+\frac{3.68-x}{84}=0,04$

$\therefore n _{ CaCO _3}=\frac{2}{100}=0.02$

and $n _{ MgCO_{3}}=\frac{1.68}{84}=0.02$

mole $\%$ of $CaCO _{3}=\frac{0.02}{0.04} \times 100=50 \%$

mole $\%$ of $MgCO _{3}=\frac{0.02}{0.04} \times 100=50 \%$