$n_{CaCO_3}=\dfrac{m}{M}=\dfrac{50}{100}=0.5(mol)$

$CaCO_3 \xrightarrow{t^o} CaO+CO_2$

$0.5$    `->`    $0.5(mol)$

`=>`$n_{CaO,lt}=0.5(mol)$

$n_{CaO,thực}=\dfrac{m}{M}=\dfrac{25}{56}(mol)$

$H\%=\dfrac{n_{CaO,thực}}{n_{CaO,lt}} \times 100\%=\dfrac{\dfrac{25}{56}}{0.5}=≈89.29\%$

`CaCO_3` $\xrightarrow[]{t^o}$ `CaO+CO_2`

`Theo\ pt:n_{CaO\ l t} = n_{CaCO_3\ l t} = (50)/(100) = 0,5\ (mol)`

`-> H = (25)/(0,5 xx 56) xx 100 ~~ 89,29\%`