\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{6,72}{22.4}=0,3\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
2.............6...............2...............3 (mol)
0,3...........................................0,3(mol)
\(a=m_{Al}=n.M=0,3.27=8,1\left(g\right)\)
\(4H_2+Fe_3O_4\rightarrow3Fe+4H_2O\)
4..............1..................3..........4 (mol)
0,3.......0,075.............0,225...... (mol)
\(n_{Fe_3O_4}=\dfrac{m}{M}=\dfrac{23,2}{232}=0,1\left(mol\right)\)
Lập tỉ lệ:\(\dfrac{0,3}{4}< \dfrac{0,1}{1}\)\(\Rightarrow\)\(Fe_3O_4\) dư, \(H_2\) hết
\(m_{Fe}=n.M=0,225.56=12,6\left(g\right)\)
\(m_{Fe_3O_4\left(du\right)}=n_{Fe_3O_4\left(du\right)}.M=\left(0,1-0,075\right).232=5,8\left(g\right)\)
\(x=m_{Fe}+m_{Fe_3O_4\left(du\right)}=12,6+5,8=21,1\left(g\right)\)
