Ta thấy Nhôm sẽ p/ứ với dd CuSO4 trước rồi mới đến Sắt
Gọi số mol p/ứ của Sắt là x (mol)
PTHH: \(2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\)
0,1______0,15______0,05_____0,15 (mol)
\(Fe+CuSO_4\rightarrow FeSO_4+Cu\)
x______x_______x______x (mol)
Ta có: \(m_{Cu}-m_{Fe\left(p/ứ\right)}-m_{Al}=7,7\)
\(\Rightarrow64\left(x+0,15\right)-0,1\cdot27-56x=7,7\) \(\Rightarrow x=0,1\)
\(\Rightarrow\left\{{}\begin{matrix}n_{FeSO_4}=0,1\left(mol\right)\\n_{Cu}=n_{CuSO_4}=0,25\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\\m_{FeSO_4}=0,1\cdot152=15,2\left(g\right)\\m_{ddCuSO_4}=\dfrac{0,25\cdot160}{20\%}=200\left(g\right)\\m_{Cu}=0,25\cdot64=16\left(g\right)\end{matrix}\right.\)
Mặt khác: \(\left\{{}\begin{matrix}m_{Al}=0,1\cdot27=2,7\left(g\right)\\m_{Fe\left(ban.đầu\right)}=0,15\cdot56=8,4\left(g\right)\\m_{Fe\left(dư\right)}= \left(0,15-0,1\right)\cdot56=2,8\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{dd\left(sau.p/ứ\right)}=m_{Al}+m_{Fe}+m_{ddCuSO_4}-m_{Cu}-m_{Fe\left(dư\right)}=192,3\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{17,1}{192,3}\cdot100\%\approx8,89\%\\C\%_{FeSO_4}=\dfrac{15,2}{192,3}\cdot100\%\approx7,9\%\end{matrix}\right.\)
