Đáp án:
$5/$
$2Al+3H_2SO_4→Al_2(SO_4)_3+3H_2$
$n_{Al}=0,2(mol)$
$n_{H_2SO_4}=0,3(mol)$
$Al$ hết, $H_2SO_4$ hết
$n_{H_2}=0,3mol$
$V=6,72l$
$m_{dd}=5,4+147-0,6=151,8g$
$C\%=\frac{34,2}{151,8}.100=2253\%$
$6/$
$a/$
$Mg+2HCl→MgCl_2+H_2$
$2Al+6HCl→2AlCl_3+3H_2$
$24a+27b=6,93
$95a+133,5=31,425$
$a=0,12; b=0,15$
$n_{H_2}=0,12+0,15.1,5=03,45mol$
$V=7,728l$
$\%m_{Mg}=\frac{2,88}{6,93}.100=41,56\%$
$\%m_{Al}=58,44\%$