\(2Fe+6H_2SO_4\left(đặc\right)\underrightarrow{to}Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\\ n_{SO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ n_{Fe}=\dfrac{2}{3}.0,25=\dfrac{1}{6}\left(mol\right)\\ \Rightarrow m=m_{Fe}=\dfrac{1}{6}.56\approx9,333\left(g\right)\)
\(n_{SO2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Pt : \(2Fe+6H_2SO_{4\left(đặc,nóng\right)}\rightarrow Fe_2\left(SO_4\right)_3+3SO_2+6H_2\)\(O\)
2 6 1 3 6
\(\dfrac{1}{6}\) 0,25
\(n_{Fe}=\dfrac{0,25.2}{3}=\dfrac{1}{6}\left(mol\right)\)
⇒ \(m_{Fe}=\dfrac{1}{6}.56=9,3\left(g\right)\)
Chúc bạn học tốt
nSO2= \(\dfrac{5,6}{64}\)=0,0875 mol
2Fe + 6H2SO4 \(\rightarrow\)Fe2(SO4)3+6H2O +3SO2
0,06 \(\leftarrow\) 0,09
mFe= 0,06 .56 =3,36 g
