hóa hợp chất vô cơ

N

nhokdangyeu01

Ta có
$2Al+Fe_2O_3-->Al_2O_3+2Fe$
mà $n_{Al}=\frac{2,7}{27}=0,1$ (mol)
$n_{Fe_2O_3}=\frac{11,2}{160}=0,07$ (mol)
\Rightarrow $Fe_2O_3$ dư, Al hết
\Rightarrow $n_{Fe_2O_{3dư}}=0,02$ mol, $n_{Fe}$=0,1 mol, $n_{Al_2O_3}=0,05$ mol
\Rightarrow $m_{Fe_2O_{3dư}}=3,2g$, $m_{Fe}=5,6g$, $m_{Al_2O_3}=5,1g$
 
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