a) Al₂O₃ + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂O

b) \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

PTHH: Al₂O₃ + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂O

                0,1---->0,3------->0,1

=> m = 0,1.342 = 34,2 (g)

c) \(C\%_{dd.H_2SO_4}=\dfrac{0,3.98}{120}.100\%=24,5\%\)

Al2O3+3H2SO4->Al2(SO4)3+3H2O

0,1------0,3------0,1-------------------0,3

n Al2O3=0,1 mol

m Al2(SO4)3=0,1.342=34,2g

C%=\(\dfrac{0,3.98}{120}100=24,5\%\)

\(a.Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2 O\\ b.n_{Al_2O_3}=0,1\left(mol\right)\\ n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=0,1\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\\ c.n_{H_2SO_4}=3n_{Al_2O_3}=0,3\left(mol\right)\\ C\%_{H_2SO_4}=\dfrac{0,3.98}{120}.100=24,5\%\)

\(n_{Al\left(OH\right)_3}=\dfrac{7,8}{78}=0,1mol\\ 2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\\ 0,1................0,15.............0,05............0,3\\ C_{\%H_2SO_4}=\dfrac{0,15.98}{300}\cdot100\%=4,9\%\\ C_{\%Al_2\left(SO_4\right)_3}=\dfrac{0,05.342}{7,8+300}\cdot100\%=5,56\%\)

a) CuO + H₂SO₄ → CuSO₄ + H₂O

b) \(n_{CuO}=\frac{16}{80}=0,2\left(mol\right)\)

\(m_{H_2SO_4}=147\times20\%=29,4\left(g\right)\)

\(n_{H_2SO_4}=\frac{29,4}{98}=0,3\left(mol\right)\)

Theo pT: \(n_{CuO}=n_{H_2SO_4}\)

Theo bài: \(n_{CuO}=\frac{2}{3}n_{H_2SO_4}\)

Vì \(\frac{2}{3}< 1\) ⇒ H₂SO₄ dư

Dung dịch sau pư gồm: H₂SO₄ dư và CuSO₄

\(m_{dd}saupư=16+147=163\left(g\right)\)

Theo PT: \(n_{H_2SO_4}pư=n_{CuO}=0,2\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4}dư=0,3-0,2=0,1\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}dư=0,1\times98=9,8\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}dư=\frac{9,8}{163}\times100\%=6,01\%\)

Theo PT: \(n_{CuSO_4}=n_{CuO}=0,2\left(mol\right)\)

\(\Rightarrow m_{CuSO_4}=0,2\times160=32\left(g\right)\)

\(\Rightarrow C\%_{CuSO_4}=\frac{32}{163}\times100\%=19,63\%\)

\(n_{CuO}=\dfrac{1.6}{80}=0.02\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{100\cdot20\%}{98}=\dfrac{10}{49}\left(mol\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(1...........1\)

\(0.02..........\dfrac{10}{49}\)

\(LTL:\dfrac{0.02}{1}< \dfrac{10}{49}\Rightarrow H_2SO_4dư\)

\(m_{\text{dung dịch sau phản ứng}}=1.6+100=101.6\left(g\right)\)

\(C\%_{CuSO_4}=\dfrac{0.02\cdot160}{101.6}\cdot100\%=3.15\%\)

\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(\dfrac{10}{49}-0.02\right)\cdot98}{101.6}\cdot100\%=17.75\%\)

Giúp mình với 

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a.Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ m_{ddHCl}=\dfrac{0,4.36,5.100}{7,3}=200\left(g\right)\\ c.m_{ddsau}=4,8+200-0,2.2=204,4\left(g\right)\\ C\%_{ddMgCl_2}=\dfrac{0,2.95}{204,4}.100\approx9,295\%\\ d.V_{ddHCl}=\dfrac{200}{1,05}=\dfrac{4000}{21}\left(ml\right)=\dfrac{4}{21}\left(l\right)\\ C_{MddHCl}=\dfrac{0,4}{\dfrac{4}{21}}=2,1\left(M\right)\)

`a)PTHH:`

  `Zn + 2HCl -> ZnCl_2 + H_2`

`0,02`                       `0,02`    `0,02`        `(mol)`

`n_[Zn]=[1,3]/65=0,02(mol)`

`b)V_[H_2]=0,02.22,4=0,448(l)`

`c)C%_[ZnCl_2]=[0,02.136]/[1,3+50-0,02.2].100~~5,31%`

\(n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,02   0,04         0,02       0,02  ( mol )

\(V_{H_2}=0,02.22,4=0,448\left(l\right)\)

\(C\%_{ZnCl_2}=\dfrac{0,02.136}{1,3+50-0,02.2}.100=5,3\%\)

\(n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) 
         0,02                 0,02           0,02 
\(V_{H_2}=0,02.22,4=0,448\left(l\right)\\ m_{\text{dd}}=1,3+50-\left(0,02.2\right)=51,26\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,02.136}{51,26}.100\%=5,3\%\)

\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)

a) Pt : \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O|\)  

             1            1                1            1

           0,2          0,2             0,2

b) \(n_{H2SO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{H2SO4}=0,2.98=19,6\left(g\right)\)

\(C_{ddH2SO4}=\dfrac{19,6.100}{200}=9,8\)⁰/₀

c) \(n_{CuSO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{CuSO4}=0,2.160=32\left(g\right)\)

\(m_{ddspu}=16+200=216\left(g\right)\)

\(C_{CuSO4}=\dfrac{32.100}{216}=14,81\)⁰/₀

\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂SO₄ --> Fe₂(SO₄)₃ + 3H₂O

______0,05------>0,15--------->0,05

=> m_H2SO4 = 0,15.98 = 14,7(g)

=> \(C\%\left(H_2SO_4\right)=\dfrac{14,7}{100}.100\%=14,7\%\)

\(C\%\left(Fe_2\left(SO_4\right)_3\right)=\dfrac{0,05.400}{8+100}.100\%=18,52\%\)

PTHH: Fe₂(SO₄)₃ + 6NaOH --> 2Fe(OH)₃\(\downarrow\) + 3Na₂SO₄

________0,05----------------------->0,1

=> mFe(OH)₃ = 0,1.107=10,7(g)

\(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{100.20\%}{98}=\dfrac{10}{49}\left(mol\right)\)

PTHH :

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

0,02       0,02           0,02       0,02 

\(\dfrac{0,02}{1}< \dfrac{\dfrac{10}{49}}{1}\) ---> H2SO4 dư và tính theo CuO 

\(C\%_{CuSO_4}=\dfrac{0,02.160}{1,6+100}.100\%\approx3,15\%\)

\(C\%_{H_2SO_4dư}=\dfrac{\left(\dfrac{10}{49}-0,02\right).98}{100+1,6}\approx17,76\left(\%\right)\)