$a,PTPƯ:$
$2Na+2H_2O\xrightarrow{} 2NaOH+H_2↑$ $(1)$
$Ca+2H_2O\xrightarrow{} Ca(OH)_2+H_2↑$ $(2)$
$n_{H_2}=\dfrac{7,84}{22,4}=0,35mol.$
$Theo$ $pt1:$ $n_{H_2}=\dfrac{1}{2}n_{Na}$
$Theo$ $pt2:$ $n_{H_2}=n_{Ca}$
Gọi $n_{Na}$ là a, $n_{Ca}$ là b.
Theo đề bài ta có hệ pt: $\left \{ {{23a+40b=14,6} \atop {0,5a+b=0,35}} \right.$ $⇒\left \{ {{a=0,2} \atop {b=0,25}} \right.$
$⇒\%m_{Na}=\dfrac{0,2.23}{14,6}.100\%=31,5\%$
$⇒\%m_{Ca}=\dfrac{0,25.40}{14,6}.100\%=68,5\%$
$b,$
$NaOH:Natri$ $hiđrôxit.$
$Ca(OH)_2:Canxi$ $hiđrôxit.$
$Theo$ $pt1:$ $n_{NaOH}=n_{Na}=0,2mol.$
$⇒m_{NaOH}=0,2.40=8g.$
$Theo$ $pt2:$ $n_{Ca(OH)_2}=n_{Ca}=0,25mol.$
$⇒m_{Ca(OH)_2}=0,25.74=18,5g$
chúc bạn học tốt!
