mH2 = 8,3 - 7,8 = 0,5 (g) => nH2 = 0,5/2 = 0,25 (mol)
PTHH :
2Al + 6HCl ---> 2AlCl3 + 3H2
x -------------> 3/2 x (mol)
Fe + 2HCl ----> FeCl2 + H2
y --------------> y (mol)
=> \(\left\{{}\begin{matrix}27x+56y=8,3\\\dfrac{3}{2}x+y=0,25\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
=> BT Al : nAlCl3 = nAl = 0,1 (mol)
BT Fe : nFeCl2 = nFe = 0,1 (mol)
=> m muối = 133,5.0,1 + 127.0,1 = 26,05 (g)
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