\(a) 2NaOH + H_2SO_4 \to Na_2SO_4 + H_2O\\ n_{NaOH} = 2n_{H_2SO_4} = 0,1.1.2 = 0,2(mol)\\ \Rightarrow V_{dd\ NaOH} = \dfrac{0,2}{1} = 0,2(lít)\\ b) Na_2SO_3 + H_2SO_4 \to Na_2SO_4 + SO_2 + H_2O\\ n_{SO_2} = n_{Na_2SO_3} = \dfrac{12,6}{126} = 0,1(mol)\\ V_{SO_2} = 0,1.22,4 = 2,24(lít)\)

Ta có: \(n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1\left(mol\right)\)

a. PTHH: Na₂SO₃ + 2HCl ---> 2NaCl + SO₂ + H₂O

Theo PT: \(n_{SO_2}=n_{Na_2CO_3}=0,1\left(mol\right)\)

=> \(V_{SO_2}=0,1.22,4=2,24\left(lít\right)\)

b. Theo PT: \(n_{HCl}=2.n_{SO_2}=2.0,1=0,2\left(mol\right)\)

=> \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

=> \(C_{\%_{HCl}}=\dfrac{7,3}{150}.100\%=4,87\%\)

c. Ta có: \(m_{dd_{NaCl}}=n_{Na_2SO_{3_{PỨ}}}=50\left(g\right)\)

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

=> \(m_{NaCl}=0,2.58,5=11,7\left(g\right)\)

=> \(C_{\%_{NaCl}}=\dfrac{11,7}{50}.100\%=23,4\%\)

n SO2 = 4,48/ 22,4 = 0,2 mol

n KOH = 0,3*1 = 0,3 mol

n SO2 < n KOH --> n KOH dư

SO2 + 2KOH --- K2SO3 + H2O

0,2   <   0,3              0,2

CM K2SO4 = 0,2/0,3 = 0,666 M

\(n_{SO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right);n_{KOH}=0,3.1=0,3\left(mol\right)\)

Ta có: \(T=\dfrac{n_{KOH}}{n_{SO_2}}=\dfrac{0,3}{0,2}=1,5\) ⇒ tạo ra muối K₂SO₃ và KHSO₃ 

PTHH: SO₂ + 2KOH → K₂SO₃ + H₂O

Mol:       x         2x              x  

PTHH: SO₂ + KOH → KHSO₃

Mol:       y          2y          y

Ta có: \(\left\{{}\begin{matrix}x+y=0,2\\2x+y=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)

\(C_{M_{ddK_2SO_3}}=\dfrac{0,1}{0,3}=0,333M\)

\(C_{M_{ddKHSO_3}}=\dfrac{0,1}{0,3}=0,333M\)

\(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)

PTHH: Mg + 2H₂SO₄ --> MgSO₄ + SO₂ + 2H₂O

          0,15---------------------------->0,15

=> \(V_{SO_2}=0,15.22,4=3,36\left(l\right)\)

bảo toàn e: nMg=nSO2=0,15 mol

=>V=0,15.22,4=3,36l

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

b, Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{20}.100\%=32,5\%\\\%m_{Al_2O_3}=67,5\%\end{matrix}\right.\)

c, Ta có: m_Al2O3 = 20 - 0,1.65 = 13,5 (g)

\(\Rightarrow n_{Al_2O_3}=\dfrac{13,5}{102}=\dfrac{9}{68}\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Zn}+6n_{Al_2O_3}=\dfrac{169}{170}\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{\dfrac{169}{170}}{1}\approx0,994\left(l\right)\)

d, Theo PT: \(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\\n_{AlCl_3}=2n_{Al_2O_3}=\dfrac{9}{34}\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,1.136=13,6\left(g\right)\\m_{AlCl_3}=\dfrac{9}{34}.133,5\approx35,34\left(g\right)\end{matrix}\right.\)

\(n_{Fe}=\dfrac{1,12}{56}=0,02\left(mol\right)\)

\(n_{H2SO4}=0,05.1=0,05\left(mol\right)\)

Pt : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

a) Xét tỉ lệ : \(0,02< 0,05\Rightarrow H2SO4dư\)

Theo Pt : \(n_{FeSO4}=n_{H2}=n_{Fe}=0,02\left(mol\right)\)

 \(\Rightarrow V_{H2\left(dktc\right)}=0,02.22,4=0,448\left(l\right)\)

b) \(n_{H2SO4\left(dư\right)}=0,05-0,02=0,03\left(mol\right)\)

\(V_{ddH2SO4\left(dư\right)}=\dfrac{0,03}{1}=0,03\left(l\right)=30\left(ml\right)\)

c) \(C_{MFeSO4}=\dfrac{0,02}{0,05}=0,4\left(M\right)\)

\(C_{MH2SO4\left(dư\right)}=\dfrac{\left(0,05-0,02\right)}{0,05}=0,6\left(M\right)\)

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