\(C1\\ n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ b,H_2+CuO\rightarrow\left(t^o\right)Cu+H_2O\\ n_{Cu\left(LT\right)}=n_{H_2}=0,3\left(mol\right)\\ n_{Cu\left(TT\right)}=n_{Cu\left(LT\right)}.H=0,3.80\%=0,24\left(mol\right)\\ m_{Cu\left(TT\right)}=0,24.80=19,2\left(g\right)\)

- Oxit:

+ Oxit bazo: Al₂O₃ (Nhôm oxit), CuO (Đồng (II) oxit)

+ Oxit axit: N₂O₅ (dinito pentaoxit)

- Axit: 

H₂CO₃ (Axit cacbonic), H₃PO₄ (axit photphoric)

- Bazo:

KOH (Kali hidroxit), Zn(OH)₂ (Kẽm hidroxit)

- Muối: 

KHSO₄ (Kali hidrosunfat), CuCl₂ (Đồng (II) clorua), ZnSO₄ (Kẽm sunfat)

a/ \(n_{KOH}=0,2.1=0,2\left(mol\right);n_{H_2SO_4}=0,3.1=0,3\left(mol\right)\)

PTHH: 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

Mol:        0,2         0,1            0,1

Ta có: \(\dfrac{0,2}{2}< \dfrac{0,3}{1}\) ⇒ KOH hết, H₂SO₄ dư

b/ \(m_{H_2SO_4dư}=\left(0,3-0,1\right).98=19,6\left(g\right)\)

c/ V_{dd sau pứ} = 0,2 + 0,3 = 0,5 (l)

d/ \(C_{M_{ddK_2SO_4}}=\dfrac{0,1}{0,5}=0,2M\)

 \(C_{M_{ddH_2SO_4dư}}=\dfrac{0,3-0,1}{0,5}=0,4M\)

mH2SO4=200.9,8%=19,6(g) -> nH2SO4=19,6/98=0,2(mol)

mKOH=200.5,6%=11,2(g) -> nKOH=11,2/56=0,2(mol)

PTHH: 2 KOH + H2SO4 -> K2SO4 + 2 H2O

Ta có: 0,2/2 < 0,2/1

=> H2SO4 dư, KOH hết, tính theo nKOH 

=> Chất có trong dd thu được sau p.ứ: K2SO4 và H2SO4(dư)

nH2SO4(p.ứ)=nK2SO4=1/2. nKOH=1/2. 0,2=0,1(mol)

nH2SO4(dư)=0,2-0,1=0,1(mol) => mH2SO4(dư)=0,1.98=9,8(g)

mK2SO4=174. 0,1=17,4(g)

mddsau= mddH2SO4 + mddKOH= 200+200=400(g)

=>C%ddH2SO4(dư)= (9,8/400).100=2,45%

C%ddK2SO4=(17,4/400).100=4,35%

\(m_{H_2SO_4}=19,6\left(g\right)\Rightarrow n_{H_2SO_4}=0,2\left(mol\right)\)

\(m_{KOH}=11,2\left(g\right)\Rightarrow n_{KOH}=0,2\left(mol\right)\)

PTHH: \(H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\)

\(\Rightarrow H_2SO_4\) dư.

\(\Rightarrow n_{H_2SO_4dư}=0,3\left(mol\right);n_{K_2SO_4}=0,1\left(mol\right)\)

\(\Rightarrow C\%_{H_2SO_4dư}=\dfrac{0,3.98}{400}.100\%=7,35\%\)

\(C\%_{K_2SO_4}=\dfrac{0,1.174}{400}.100\%=4,35\%\)

\(m_{H_2SO_4}=\dfrac{19,6.100}{100}=19,6\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)

PTHH: 2KOH + H₂SO₄ ---> K₂SO₄ + 2H₂O

             0,4<-----0,2--------->0,2

\(\rightarrow m_{ddKOH}=\dfrac{0,4.56}{5,6\%}=400\left(g\right)\\ m_{dd\left(sau.pư\right)}=400+100=500\left(g\right)\\ m_{K_2SO_4}=174.0,2=34,8\left(g\right)\\ \rightarrow C\%_{K_2SO_4}=\dfrac{34,8}{500}.100\%=6,96\%\)

\(n_{H_2SO_4}=\dfrac{100.19,6\%}{98}=0,2mol\)

\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

  0,4            0,2              0,2                  ( mol )

\(m_{ddKOH}=\dfrac{0,4.56}{5,6\%}=400g\)

\(C\%_{K_2SO_4}=\dfrac{0,2.174}{100+400}.100=6,96\%\)

\(n_{H_2SO_4}=\dfrac{4,9}{98}=0,05\left(mol\right)\\ H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\\ a,n_{NaOH}=n_{H_2O}2.0,05=0,1\left(mol\right)\\ \Rightarrow m_{NaOH}=0,1.40=4\left(g\right)\\ b,C1:n_{Na_2SO_4}=n_{H_2SO_4}=0,05\left(mol\right)\\ m_{sp}=m_{Na_2SO_4}+m_{H_2O}=0,05.142+18.0,1=8,9\left(g\right)\\ C2:m_{sp}=m_{H_2SO_4}+m_{NaOH}=4,9+4=8,9\left(g\right)\)