Ta có nH2 = \(\dfrac{3,36}{22,4}\) = 0,15 ( mol )
Fe + 2HCl \(\rightarrow\) FeCl2 + H2
x........2x..............x.........x
Zn + 2HCl \(\rightarrow\) ZnCl2 + H2
y.......2y................y........y
=> \(\left\{{}\begin{matrix}56x+65y=9,3\\x+y=0,15\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)
=> VHCl = n : CM = ( 2 . 0,05 + 2 . 0,1 ) : 1 = 0,3 ( lít )
=> mFe = 56 . 0,05 = 2,8 ( gam )
=> %mFe = \(\dfrac{2,8}{9,3}\times100\approx30\%\)
=> %mZn = 100 -30 = 70 %
=> mFeCl2 = 127 . 0,05 = 6,35 ( gam )
=> mZnCl2 = 136 . 0,1 = 13,6 ( gam )
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