$n_{H_2SO_4}=0,2x (mol)$
$n_{KOH}=0,3y (mol)$
$2KOH+H_2SO_4\to K_2SO_4+2H_2O$
$\Rightarrow n_{KOH\text{dư}}=-0,4x+0,3y(mol)$
$100ml$ E cần $0,04$ mol $H_2SO_4$
$\Rightarrow$ $500ml$ E cần $0,2$ mol $H_2SO_4$
$\Rightarrow -0,4x+0,3y=0,2.2=0,4$ (1)
$n_{H_2SO_4}=0,3x (mol)$
$n_{KOH}=0,2y (mol)$
$n_{Al}=0,04(mol)$
* Nếu dư $H_2SO_4$
$\Rightarrow n_{H_2SO_4\text{dư}}=0,3x-0,1y(mol)$
$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$
$\Rightarrow 0,3x-0,1y=0,04.1,5=0,06$ (2)
$(1)(2)\to x=1,16; y=2,88$
* Nếu dư $KOH$
$\Rightarrow n_{KOH\text{dư}}=-0,6x+0,2y (mol)$
$2Al+2KOH+2H_2O\to 2KAlO_2+3H_2$
$\Rightarrow -0,6x+0,2y=0,04$ (3)
$(1)(3)\to x=0,68; y=2,24$