\(2Fe+6H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)
Gọi 0,375a là mol Fe \(\Rightarrow\) nH2SO4= a mol
\(n_{Fe_{pu}}=\frac{a}{3}\left(mol\right)\)
Nên dư a/24 mol Fe. Tạo a/6 mol Fe2(SO4)3
\(Fe+Fe_2\left(SO_4\right)_3\rightarrow3FeSO_4\)
\(\Rightarrow\) nFeSO4= a/8 mol. Dư a/8 mol Fe2(SO4)3.
m muối= 8,28g
\(\Rightarrow\frac{152a}{8}+\frac{400a}{8}=8,28\)
\(\Rightarrow a=0,12\)
nFe phản ứng= 0,375a= 0,045 mol
\(\Rightarrow m_{Fe}=2,52\left(g\right)\)
\(n_{H2SO4}=0,12\left(mol\right)\Rightarrow n_{SO2}=0,06\left(mol\right)\)
\(n_{NaOH}=0,1\left(mol\right)\)
\(\frac{n_{NaOH}}{n_{SO2}}=1,67\Rightarrow\) Tạo 2 muối
\(NaOH+SO_2\rightarrow NaHSO_3\)
\(2NaOH+SO_2\rightarrow Na_2SO_3+H_2O\)
Gọi x là mol NaHSO3, y là mol Na2SO3
\(\left\{{}\begin{matrix}x+2y=0,1\\x+y=0,06\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,04\end{matrix}\right.\)
\(CM_{NaHSO3}=\frac{0,02}{0,1}=0,2M\)
\(CM_{Na2SO3}=\frac{0,04}{0,1}=0,4M\)
