Đặt nAl = x(mol); nMg = y (mol); nFe = z (mol); ( x, y, z > 0 )
nH2 = 0,6 mol
2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2 (1)
Mg + 2HCl \(\rightarrow\) MgCl2 + H2 (2)
Fe + 2HCl \(\rightarrow\) FeCl2 + H2 (3)
\(n_{Mg}=n_{Al}\)
\(\Leftrightarrow y-x=0\) (4)
Từ (1)(2)(3)(4) ta có hệ pt
\(\left\{{}\begin{matrix}27x+24y+56z=15,8\\1,5x+y+z=0,6\\y-x=0\end{matrix}\right.\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}x=0,2\\y=0,2\\z=0,1\end{matrix}\right.\)
\(\Rightarrow\) %Al = \(\dfrac{0,2.27.100}{15,8}\)\(\approx\) 34,2%
\(\Rightarrow\) %Mg = \(\dfrac{0,2.24.100}{15,8}\) \(\approx\) 30,4%
\(\Rightarrow\) %Fe = \(\dfrac{0,1.56.100}{15,8}\) \(\approx\) 35,4%
Đúng 0
Bình luận (0)
