$a,PTPƯ:$
$Mg+2CH_3COOH\xrightarrow{} (CH_3COO)_2Mg+H_2↑$ $(1)$
$CaO+2CH_3COOH\xrightarrow{} (CH_3COO)_2Ca+H_2O$ $(2)$
$n_{H_2}=\dfrac{4,48}{22,4}=0,2mol.$
$Theo$ $pt1:$ $n_{Mg}=n_{H_2}=0,2mol.$
$⇒m_{Mg}=0,2.24=4,8g.$
$⇒m_{CaO}=15-4,8=10,2g.$
$b,\%m_{Mg}=\dfrac{4,8}{15}.100\%=32\%$
$\%m_{CaO}=\dfrac{10,2}{15}.100\%=68\%$
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