$n_{H_2}=8,96/22,4=0,4mol$
$Fe+2HCl\to FeCl_2+H_2↑(1)$
$2Al+6HCl\to 2AlCl_3+3H_2↑(2)$
a.Gọi $n_{Fe}=a,n_{Al}=b(a,b>0)$
Ta có :
$m_{hh}=56a+27b=11g$
$n_{H_2}=a+1,5b=0,4mol$
Ta có hpt :
$\left\{\begin{matrix} 56a+27b=11 & \\ a+1,5b=0,4 & \end{matrix}\right.$ $⇔\left\{\begin{matrix} a=0,1 & \\ b=0,2 & \end{matrix}\right.$
$⇒\%m_{Fe}=\dfrac{56.0,1}{11}.100\%=50,9\%$
$\%m_{Al}=100\%-50,9\%=49,1\%$
b.Theo pt (1) và (2) :
$n_{FeCl_2}=n_{Fe}=0,1mol⇒m_{FeCl_2}=0,1.127=12,7g$
$n_{AlCl_3}=n_{Al}=0,2mol⇒m_{AlCl_3}=0,1.133,5=13,35g$
$m_{dd\ spu}=11+500-0,4.2=510,2g$
$⇒C\%_{FeCl_2}=\dfrac{12,7}{510,2}.100\%=2,5\%$
$C\%_{AlCl_3}=\dfrac{13,35}{510,2}.100\%=2,61\%$
c.Theo pt (1) và (2) :
$n_{HCl}=n_{HCl(1)}+n_{HCl(2)}=0,1.2+0,2.3=0,8mol$
$⇒m_{HCl}=0,8.36,5=29,2g$
$⇒C\%_{HCl}=\dfrac{29,2}{500}.100\%=5,84\%$
$d.2H_2+O_2\overset{t^o}\to 2H_2O$
Theo pt :
$n_{O_2}=1/2.n_{H_2}=1/2.0,4=0,2mol$
$⇒V_{O_2}=0,2.22,4=4,48l$
$⇒V_{kk}=4,48.5=22,4l$