\(m_{Al}=19,2\%.27,8=5,3376\left(g\right)\Rightarrow n_{Al}=0,2\left(mol\right)\)
\(m_{Fe}=27,8-5,3376=22,4624\left(g\right)\Rightarrow n_{Fe}=0,4\left(mol\right)\)
\(4Al+3O_2-^{t^o}\rightarrow2Al_2O_3\)
\(3Fe+2O_2-^{t^o}\rightarrow Fe_3O_4\)
Theo PT : \(n_{O_2}=0,2.\dfrac{3}{4}+0,4.\dfrac{2}{3}=\dfrac{5}{12}\left(mol\right)\)
Vì oxi chiếm 20% thể tích không khí
=> \(V_{kk}=\dfrac{5}{12}.22,4.\dfrac{100}{20}=\dfrac{140}{3}\left(lít\right)=46,67\left(lít\right)\)
Bảo toàn khối lượng ta có: \(m_{KL}+m_{O_2}=m_{oxit}\)
=> \(m_{oxit}=27,8+\dfrac{5}{12}.32=\dfrac{617}{15}\left(g\right)=41,13\left(g\right)\)
a,
mAl=27,8.19,42%=5,4gmAl=27,8.19,42%=5,4g
⇒nAl=5,427=0,2mol⇒nAl=5,427=0,2mol
⇒nFe=27,8−5,456=0,4mol⇒nFe=27,8−5,456=0,4mol
4Al+3O2to→2Al2O34Al+3O2→to2Al2O3
3Fe+2O2to→Fe3O43Fe+2O2→toFe3O4
⇒nO2=34nAl+23nFe=512mol⇒nO2=34nAl+23nFe=512mol
⇒Vkk=512.22,4.5=46,67l⇒Vkk=512.22,4.5=46,67l
b,
mrắn=27,8+mO2=27,8+512.32=41,1g
\(m_{Al}=27,8.\dfrac{19,2}{100}=5,4\) ( g )
\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\) ( mol )
\(m_{Fe}=27,8-5,4=22,4\) ( g )
\(n_{Fe}=\dfrac{m}{m}=\dfrac{22,4}{56}=0,4\) ( mol )
\(4Al\) + \(3O_2\) → \(2Al_2O_3\)
\(3Fe\) + \(2O_2\) → \(Fe_3O_4\)
\(n_{O_2}=\dfrac{3}{4}Al+\dfrac{2}{3}Fe=\dfrac{3}{4}.0,2+\dfrac{2}{3}.0,4=\dfrac{5}{12}\) ( mol )
\(V_{kk}=5V_{O_2}=5.\dfrac{5}{12}.22,4\approx46,7\) ( \(l\) )
\(4Al\) + \(3O_2\) → \(2Al_2O_3\)0,2 mol → 0,1 mol
\(3Fe\) + \(2O_2\) → \(Fe_3O_4\)0,4 mol → \(\dfrac{0,4}{3}\) mol\(m_{Al_2O_3}=n.M=0,1.102=10,2\) ( g ) \(m_{Fe_3O_4}=n.M=\dfrac{0,4}{3}.232\approx31\) ( g )⇒ \(m_{ran}=10,2+31=316,2\) ( g )
