2Al + 6HCl → 2AlCl3 + 3H2
x \(\dfrac{{3y}}{{2}}\) (mol)
Mg + 2HCl → MgCl2 + H2
y y (mol)
Ta có hệ phương trình :
27x + 24y = 1,5
\(\dfrac{{3y}}{{2}}\) + y = \(\dfrac{{1,68}}{{22,4}} = 0,075\)
=> \(x= \dfrac{{1}}{{30}}, \;y=0,025\)
\(m_{Al}=27\times \dfrac{{1}}{{30}}=0,9 \;g\)
%\(Al=\dfrac{{0,9}}{{1,5}}\times 100\)%=60%
=>%Mg= 40%