nCu = 0,3 mol
Đặt nMg = x ( mol ); nFe = y ( mol ) ( x,y > 0 )
Mg + CuSO4 \(\rightarrow\) MgSO4 + Cu (1)
x............................................x
Fe + CuSO4 \(\rightarrow\) FeSO4 + Cu (2)
y........................................y
nMg = nFe
\(\Leftrightarrow x=y\)
\(\Leftrightarrow\) x - y = 0 (3)
Từ (1)(2)(3) ta có hệ pt
\(\left\{{}\begin{matrix}x+y=0,3\\x-y=0\end{matrix}\right.\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}x=0,15\\y=0,15\end{matrix}\right.\)
\(\Rightarrow\) m = 0,15. ( 24 + 56 ) = 12 (g)
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