$n_{H_2}=7,84/22,4=0,35mol$

$PTHH :$

$2Na+2H_2O\to 2NaOH+H_2↑(1)$

$Ca+2H_2O\to Ca(OH)_2+H_2↑(2)$

Gọi $n_{Na}=a;n_{Ca}=b(a,b>0)$

Ta có :

$m_{hh}=23a+40b=14,6g$

$n_{H_2}=0,5a+b=0,35mol$

Ta có hpt :

$\left\{\begin{matrix} 23a+40b=14,6 & \\  0,5a+b=0,35 &  \end{matrix}\right.$ $⇔\left\{\begin{matrix} a=0,2 & \\  b=0,25 &  \end{matrix}\right.$

$⇒\%m_{Na}=\dfrac{0,2.23}{14.6}.100\%=31,5\%$

$\%m_{Ca}=100\%-31,5\%=68,5\%$

b.Theo pt (1) :

$n_{NaOH}=n_{Na}=0,2mol$

$⇒m_{NaOH}=0,2.40=8g$

NaOH : natri hidroxit

Theo pt (2) :

$n_{Ca(OH)_2}=n_{Ca}=0,25mol$

$⇒m_{Ca(OH)_2}=0,25.74=18,5g$

CaOH : canxi hidroxit

 

$a,PTPƯ:$

$2Na+2H_2O\xrightarrow{} 2NaOH+H_2↑$  $(1)$

$Ca+2H_2O\xrightarrow{} Ca(OH)_2+H_2↑$  $(2)$

$n_{H_2}=\dfrac{7,84}{22,4}=0,35mol.$

$Theo$ $pt1:$ $n_{H_2}=\dfrac{1}{2}n_{Na}$

$Theo$ $pt2:$ $n_{H_2}=n_{Ca}$

Gọi $n_{Na}$ là a, $n_{Ca}$ là b.

Theo đề bài ta có hệ pt: $\left \{ {{23a+40b=14,6} \atop {0,5a+b=0,35}} \right.$ $⇒\left \{ {{a=0,2} \atop {b=0,25}} \right.$ 

$⇒\%m_{Na}=\dfrac{0,2.23}{14,6}.100\%=31,5\%$

$⇒\%m_{Ca}=\dfrac{0,25.40}{14,6}.100\%=68,5\%$

$b,$

$NaOH:Natri$ $hiđrôxit.$

$Ca(OH)_2:Canxi$ $hiđrôxit.$

$Theo$ $pt1:$ $n_{NaOH}=n_{Na}=0,2mol.$

$⇒m_{NaOH}=0,2.40=8g.$

$Theo$ $pt2:$ $n_{Ca(OH)_2}=n_{Ca}=0,25mol.$

$⇒m_{Ca(OH)_2}=0,25.74=18,5g$

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