a.b.\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,05 > 0,1 ( mol )
\(\dfrac{1}{30}\) 0,1 \(\dfrac{1}{15}\) ( mol )
\(m_{thu.được}=m_{Fe_2O_3\left(dư\right)}+m_{Fe}\)
\(=\left(0,05-\dfrac{1}{30}\right).160+\dfrac{1}{15}.56=6,4\left(g\right)\)
c. Để khử hết oxit sắt thì thể tích H2 cần dùng
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,05 0,15 ( mol )
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)