\(m_{Fe}=70\%.m_{hh}=70\%.32=22,4\left(g\right)\)
=> \(n_{Fe}=\frac{m_{Fe}}{M_{Fe}}=\frac{22,4}{56}=0,4\left(mol\right)\)
Mà \(m_{hh}=m_{Fe}+m_{Mg}=32\left(g\right)\)
-> \(m_{Mg}=m_{hh}-m_{Fe}=32-22,4=9,6\left(g\right)\)
=> \(n_{Mg}=\frac{m_{Mg}}{M_{Mg}}=\frac{9,6}{24}=0,4\left(mol\right)\)
PTHH (I) : \(2Mg+O_2\rightarrow2MgO\)
PTHH (II) : \(3Fe+2O_2\rightarrow Fe_3O_4\)
- Theo PTHH (I) : \(n_{O2}=\frac{1}{2}.n_{Mg}=\frac{1}{2}.0,4=0,2\left(mol\right)\)
- Theo PTHH (II) : \(n_{O2}=\frac{2}{3}.n_{Fe}=\frac{2}{3}.0,4=\frac{4}{15}\left(mol\right)\)
Mà \(n_{O2}=n_{O2\left(I\right)}+n_{O2\left(II\right)}=0,2+\frac{4}{15}=\frac{7}{15}\left(mol\right)\)
-> \(V_{O2}=n_{O2}.22,4=\frac{7}{15}.22,4\approx10,5\left(l\right)\)
- Mà thể tích oxit chiếm 20% thể tích không khí .
-> \(V_{kk}=5.V_{O2}=5.10,5\approx52,3\left(l\right)\)
b, - Theo PTHH (I) : \(n_{MgO}=n_{Mg}=0,4\left(mol\right)\)
-> \(m_{MgO}=n_{MgO}.M_{MgO}=0,4.\left(24+16\right)=16\left(g\right)\)
- Theo PTHH (II) : \(n_{Fe3O4}=\frac{1}{3}n_{Fe}=\frac{1}{3}.0,4\approx0,13\left(mol\right)\)
-> \(m_{Fe3O4}=n_{Fe3O4}.M_{Fe3O4}=0,13.232=30,16\left(g\right)\)
Mà mHỗn hợp tạo thành = mMgO + mFe3O4 = 16 + 30,16 = 46,16 ( g )