a) n H2=1,68/22,4 = 0,075(mol)
2Al+ 6HCl----->2AlCl3+3H2
x---------3x-----------------1,5x
Mg+2HCl--------->MgCl2+H2
y------2y---------------------y
Theo bài ta có hpt
\(\left\{{}\begin{matrix}27x+24=1,5\\1,5x+y=0,075\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,03333\\y=0,025\end{matrix}\right.\)
m Al =0,03333.27=0,9(g)
m Mg= 1,5-0,9=0,6(g)
b) Theo pthh
n HCl =2n H2=0,15(mol)
m HCl =0,15 .36,5= 5,475(g)
m dd HCl =\(\frac{5,475.100}{7,3}=75\left(g\right)\)
V HCl =\(\frac{75}{1,2}=62,5\left(ml\right)\)
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