\(Fe_2O_3+2Al\rightarrow\left(t^o\right)Al_2O_3+2Fe\\ n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
TH1: Nếu p.ứ vừa đủ thì ta có:
\(hh.rắn.X:0,1\left(mol\right)Al_2O_3.và.0,2\left(mol\right)Fe\\ m_X=0,1.10,2+0,2.56=21,4\left(g\right)< 24,1\left(g\right)\)
=> P.ứ có dư. Nên loại TH1
TH2: Khi phản ứng có dư. Nếu dư Al.
\(m_{Al\left(dư\right)}=m-0,2.27=m-5,4\left(g\right)\\m_{Fe}=0,1.2.56=11,2\left(g\right)\\ m_{Al_2O_3}=0,1.102=10,2\left(g\right)\\ m_X=\left(m-5,4\right)+11,2+10,2=24,1\\ \Leftrightarrow m=8,1\left(g\right)\\ \Rightarrow n_{Al\left(dư\right)}=\dfrac{8,1}{27}-0,1=0,2\left(mol\right)\\ -VớiTH2:\\ 2Al_{dư}+6HCl\rightarrow2AlCl_3+3H_2\\ Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{HCl\left(tổng\right)}=0,2.3+0,1.6+0,2.2=1,6\left(mol\right)\\V_{ddHCl}=\dfrac{1,6}{1}=1,6\left(lít\right)=V \)
- TH3: Nếu Al hết, Fe2O3 dư
\(X.có:\dfrac{56m}{27}\left(g\right)Fe;\dfrac{17m}{9}\left(g\right)Al_2O_3;\left(16-\dfrac{80m}{27}\right)\left(g\right)Fe_2O_3\left(dư\right)\\ m_X=24,1=\dfrac{56m}{27}+\dfrac{17m}{9}+\left(16-\dfrac{80m}{27}\right)\\ \Leftrightarrow m=8,1\left(g\right)\\ n_{Al\left(bđ\right)}=0,3\left(mol\right)>n_{Fe_2O_3}\left(LOẠI\right)\)
Vậy: V=1,6(lít)
