Since, $MgCO _{3} \xrightarrow{\Delta} MgO + CO _{2(g)}$
Mole ratio $1: 1: 1$
Mass ratio (ing) $84: 40: 44$
Given values $10\, g\, 4g \,0.1\, mol$
$(=4.4 \,g )\left\{\because n=\frac{W}{m}\right\}$
(i) $\because 84\, g .$ of $MgCO _{3}$ give $MgO =40 \,g$
$\therefore 10 \,g$ of $MgCO _{3}$ give $MgO =\frac{40 \times 10}{84}=4.76\, g$
Thus, $MgO$ obtained is less by
$4.76-4.0=0.76 \,g$
(ii) $\because 84\, g$ of $MgCO _{3}$ give $CO _{2}=44 \,g$
$\therefore 10\, g $ of $MgCO _{3}$ give $CO _{2}=\frac{44 \times 10}{84}=5.23 \,g$
Thus $CO _{2}$ obtained is less by $5 \cdot 23-4 \cdot 4=0 \cdot 84\, g$
Total short $=0 \cdot 76+0 \cdot 84=1 \cdot 59=1.60$
$\because 10 \,g .$ of $MgCO _{3}$ give less products by $=1.60\, g$
$\therefore 100 \,g$ of $MgCO _{3}$ give less products $by =1.600\, g$
Hence, $MgCO _{3}=100-16=84 \%$