$n_{H_2}=17,92/22,4=0,8mol$
$a.Mg+H_2SO_4\to MgSO_4+H_2↑(1)$
$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2↑(2)$
$Gọi\ n_{Mg}=a;n_{Al}=b (a,b>0)$
Ta có :
$m_{hh}=24a+27b=15,6g$
$n_{H_2}=a+1,5b=0,8mol$
Ta có hpt :
$\left\{\begin{matrix}
24a+27b=15,6 & \\
a+1,5b=0,8 &
\end{matrix}\right.$
$⇔\left\{\begin{matrix}
a=0,2 & \\
b=0,4 &
\end{matrix}\right.$
$⇒\%m_{Mg}=\dfrac{0,2.24}{15,6}.100\%≈30,77\%$
$\%m_{Al}=100\%-30,77\%=69,23\%$
$b.PTHH :$
$MgSO_4+Ba(OH)_2\to Mg(OH)_2+BaSO_4↓(3)$
$Al_2(SO_4)_3+3Ba(OH)_2\to 2Al(OH)_3↓+3BaSO_4↓(4)$
Theo pt (1) và (2) :
$n_{MgSO_4}=n_{Mg}=0,2mol$
$n_{Al_2(SO_4)_3}=1/2.n_{Al}=1/2.0,4=0,2mol$
Theo pt (3) và (4) :
$n_{BaSO_4(3)}=n_{MgSO_4}=0,2mol$
$⇒m_{BaSO_4(3)}=0,2.233=46,6g$
$n_{Al(OH)_3}=2.n_{Al_2(SO_4)_3}=2.0,2=0,4mol$
$⇒m_{Al(OH)_3}=0,4.78=31,2g$
$n_{BaSO_4}=3.n_{Al_2(SO_4)_3}=3.0,2=0,6mol$
$⇒m_{BaSO_4(3)}=0,6.233=139,8g$
$⇒m_{kết\ tủa}=46,6+31,2+139,8=217,6g$