\(n_{Cl_2}\)\(\dfrac{6,72}{22,4}=\) 0,3 (mol)
Gọi nFe = x (mol), nZn = y (mol) trong 15,35g hỗn hợp
2Fe + 3Cl2 \(\underrightarrow{t^o}\) 2FeCl3
x 1,5x x (mol)
Zn + Cl2 \(\underrightarrow{t^o}\) ZnCl2
y y y
\(\Rightarrow Ta\) có hệ \(\left\{{}\begin{matrix}56x+65y=15,35\\1,5x+y=0,3\end{matrix}\right.\)\(\Rightarrow\)\(\left\{{}\begin{matrix}x=0,1\\y=0,15\end{matrix}\right.\)
\(\Rightarrow\)%(m) Fe =\(\dfrac{56.0,1}{15,35}\). 100%= 36,48%
%(m) Zn = 100 - 36,48 = 63,52 %