A mixture of $CaC{O}_3$ and $MgC{O}_3$ weighing 1.84 g in heating left a residue weighing 0.96 g. Calculate the percentage of limestone in the mixture. 

${\text{CaCO}}_3+{\text{MgCO}}_3\to \text{CaO}+\text{MgO}+2{\text{CO}}_{\text{2}}$

0.01mol    0.01mol       0.01mol        0.01mol

0.01mol $CaO$ = 0.56g

0.01mol $MgO$ = 0.40g

Total residue = 0.96g

$\therefore {\text{ Mass of CaCO}}_{\text{3}}\text{\hspace{0.17em}\hspace{0.17em}in\hspace{0.17em}\hspace{0.17em}mixture\hspace{0.17em}\hspace{0.17em}}=1.0\text{g}$

Mass of  $MgC{O}_3$ in mixture = 0.84g

$\%$of $CaC{O}_3$= 1/1.84 x100 = 54