Ta có: \(n_{SO_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)

\(m_{H_2SO_4}=300.78,4\%=235,2\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{235,2}{98}=2,4\left(mol\right)\)

PT: \(2Fe+6H_2SO_{4\left(đ\right)}\underrightarrow{t^o}Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)

____0,5____1,5________0,25______0,75 (mol)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=2,4-1,5=0,9\left(mol\right)\)

Ta có: m _{dd sau pư} = m_Fe + m _{dd H2SO4} - m_SO2 

                             = 0,5.56 + 300 - 0,75.64 = 280 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,9.98}{280}.100\%=31,5\%\\C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,25.400}{280}.100\%\approx35,7\%\end{matrix}\right.\)

Bạn tham khảo nhé!

\(Fe_2O_3=\dfrac{24}{160}=0,15\left(mol\right)\\ PTHH:Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,15\left(mol\right)\\ a,m_{Fe_2\left(SO_4\right)_3}=400.0,15=60\left(g\right)\\ b,n_{H_2SO_4}=3n_{Fe_2O_3}=3.0,15=0,45\left(mol\right)\\ C_{MddH_2SO_4}=\dfrac{0,45}{0,2}=2,25\left(M\right)\\ c,V_{ddsau}=V_{ddH_2SO_4}=0,2\left(l\right)\\ C_{MddFe_2\left(SO_4\right)_3}=\dfrac{0,15}{0,2}=0,75\left(M\right)\)

\(n_{H_2SO_4}=\dfrac{500.4,9\%}{98}=0,25\left(mol\right)\)

\(n_{Fe_2\left(SO_4\right)_3}=\dfrac{500.20\%}{400}=0,25\left(mol\right)\)

Gọi số mol Na là k (mol)

\(n_{Fe\left(OH\right)_3}=\dfrac{32,1}{107}=0,3\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

             k---------------->k------>0,5k

            2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

              0,5<-----0,25------->0,25

             6NaOH + Fe₂(SO₄)₃ --> 3Na₂SO₄ + 2Fe(OH)₃

                 0,9<------0,15<---------0,45<--------0,3

=> k = 0,5 + 0,9 = 1,4 (mol)

=> m = 1,4.23 = 32,2 (g)

m_{dd sau pư} = 32,2 + 500 - 0,7.2 - 32,1 = 498,7 (g)

\(\left\{{}\begin{matrix}n_{Na_2SO_4}=0,7\left(mol\right)\\n_{Fe_2\left(SO_4\right)_3}=0,1\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}C\%_{Na_2SO_4}=\dfrac{0,7.142}{498,7}.100\%=19,93\%\\C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400}{498,7}.100\%=8,02\%\end{matrix}\right.\)

Phần 1:

Fe + 2HCl → FeCl₂  +  H₂

nH₂ = 4,48:22,4 = 0,2 mol  = nFe 

<=> mFe = 0,2.56 = 11,2 gam

=> mCu = 35,2:2 - 11,2 = 0,1 mol

Phần 2:

2Fe  +   6H₂SO₄  → Fe₂(SO₄)₃   +  3SO₂   + 6H₂O

Cu  + 2H₂SO₄  → CuSO₄   +  SO₂   + 2H₂O

=> nSO₂ = 0,3 + 0,1 = 0,4 mol <=> V = 0,4.22,4 = 8,96 lít

=>nH₂SO₄ phản ứng = 3.0,2 + 2.0,1 = 0,8 mol

=> nH₂SO₄ ban đầu = 0,8 + 0,8.15% = 0,92 mol

<=>m dung dịch = m(Fe + Cu) + mH₂SO₄ - mSO₂ = 35,2:2 + 0,92.98 - 0,4.64 = 82,16 gam

C%Fe₂(SO₄)₃ = \(\dfrac{0,1.400}{82,16}.100\%\)= 48,7%

C%CuSO₄ = \(\dfrac{0,1.160}{82,16}.100\%\) = 19,5%

C%H₂SO₄ dư = \(\dfrac{0,8.15\%.98}{82,16}\)_.100% = 14,7%