Giả sử có 1 mol Fe phản ứng
PTHH: \(2Fe+6H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)
1---->3---------------->0,5------------->1,5
`=>` \(m_{Fe_2\left(SO_4\right)_3}=0,5.400=200\left(g\right)\)
Ta có: \(C\%_{Fe_2\left(SO_4\right)_3}=C\%_{H_2SO_4\left(dư\right)}\)
`=>` \(m_{H_2SO_4\left(dư\right)}=m_{Fe_2\left(SO_4\right)_3}=200\left(g\right)\)
`=>` \(n_{H_2SO_4\left(dư\right)}=\dfrac{200}{98}=\dfrac{100}{49}\left(mol\right)\)
`=>` \(m_{ddH_2SO_4}=\dfrac{\left(\dfrac{100}{49}+3\right).98}{78,4\%}=\dfrac{30875}{49}\left(g\right)\)
`=>` \(m_{ddspư}=\dfrac{30875}{49}+56-1,5.64=\dfrac{28915}{49}\left(g\right)\)
`=>` \(C\%_{Fe_2\left(SO_4\right)_3}=C\%_{H_2SO_{4\left(dư\right)}}=\dfrac{200}{\dfrac{28915}{49}}.100\%=33,7\%\)