$a,PTPƯ:Ca+2H_2O\xrightarrow{} Ca(OH)_2+H_2↑$
$n_{Ca}=\dfrac{4}{40}=0,1mol.$
$Theo$ $pt:$ $n_{Ca(OH)_2}=n_{H_2}=n_{Ca}=0,1mol.$
$⇒m_{dd\ spư}=m_{Ca}+m_{H_2O}-m_{H_2}=4+100-(0,1.2)=103,6g.$
$⇒C\%_{Ca(OH)_2}=\dfrac{0,1.74}{103,6}.100\%=7,14\%$
$b,PTPƯ:Ca+H_2SO_4\xrightarrow{} CaSO_4+H_2↑$
$Theo$ $pt:$ $n_{H_2SO_4}=n_{Ca}=0,1mol.$
$⇒m_{H_2SO_4}=0,1.98=9,8g.$
$⇒m_{ddH_2SO_4}=\dfrac{9,8.100}{40}=24,5g.$
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