Số mol SO₂ và NaOH lần lượt là 0,2 và 0,25.

1 < OH^-/SO₂=1,25 < 2 ⇒ Dung dịch X chứa hai muối Na₂SO₃ và NaHSO₃.

\(n_{Na_2SO_3}=0,25-0,2=0,05\left(mol\right)\) ⇒ \(n_{NaHSO_3}=0,2-0,05=0,15\left(mol\right)\).

1. Khối lượng muối có trong X:

m=0,05.126+0,15.104=21,9 (g).

2. Nồng độ mol/l các chất trong X:

\(C_{M\left(Na_2SO_3\right)}\)=0,05/0,2=0,25 (mol/l).

\(C_{M\left(NaHSO_3\right)}\)=0,15/0,2=0,75 (mol/l).

3. Khối lượng kết tủa BaSO₃ là:

m'=0,2.217=43,4 (g).

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Ta có: \(n_{HCl}=\dfrac{200}{1000}.2=0,4\left(mol\right)\)

\(PTHH:Mg+2HCl--->MgCl_2+H_2\uparrow\left(1\right)\)

a. Theo PT₍₁₎: \(n_{Mg}=n_{H_2}=n_{MgCl_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,2.24=4,8\left(g\right)\\V_{H_2}=0,2.22,4=4,48\left(lít\right)\end{matrix}\right.\)

b. \(PTHH:2NaOH+MgCl_2--->Mg\left(OH\right)_2\downarrow+2NaCl\left(2\right)\)

Ta có: \(n_{NaOH}=\dfrac{\dfrac{20\%.100}{100\%}}{40}=0,5\left(mol\right)\)

Ta thấy: \(\dfrac{0,5}{2}>\dfrac{0,2}{1}\)

Vậy NaOH dư.

Theo PT₍₂₎: \(n_{Mg\left(OH\right)_2}=n_{MgCl_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Mg\left(OH\right)_2}=0,2.58=11,6\left(g\right)\)

a: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

200ml=0,2 lít

\(n_{HCl}=0.2\cdot22.4=4.48\left(mol\right)\)

\(\Leftrightarrow n_{H_2}=2.24\left(mol\right)\)

\(\Leftrightarrow m_{H_2}=n_{H_2}\cdot M=2.24\cdot1=2.24\left(g\right)\)

\(n_{MgCl_2}=2.24\left(mol\right)\)

\(\Leftrightarrow n_{Mg}=2.24\left(mol\right)\)

\(\Leftrightarrow m_{Mg}=2.24\cdot24=53.76\left(g\right)\)

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

\(SO_2+Ba\left(OH\right)_2\rightarrow BaSO_3\downarrow+H_2O\)

0,2------------------>0,2

\(m_{kt}=m_{BaSO_3}=0,2.217=43,4\left(g\right)\)

\(a,PTHH:3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\\ 2Fe\left(OH\right)_3\rightarrow^{t^o}Fe_2O_3+3H_2O\uparrow\\ b,n_{FeCl_3}=1,5\cdot0,2=0,3\left(mol\right)\\ \Rightarrow n_{NaOH}=3n_{FeCl_3}=0,9\left(mol\right)\\ \Rightarrow V_{dd_{NaOH}}=\dfrac{0,9}{2}=0,45\left(l\right)\)

Theo đề: \(\left\{{}\begin{matrix}X:Fe\left(OH\right)_3\\A:NaCl\\Y:Fe_2O_3\end{matrix}\right.\)

Theo PT: \(n_{NaCl}=3n_{FeCl_3}=0,9\left(mol\right)\)

\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,9}{0,45+0,2}\approx1,4M\)

\(c,\) Theo PT: \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,3\left(mol\right);n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_X=m_{Fe\left(OH\right)_3}=0,3\cdot107=32,1\left(g\right)\\m_Y=m_{Fe_2O_3}=0,15\cdot160=24\left(g\right)\end{matrix}\right.\)

Ta có: \(n_{SO_2}=\dfrac{2,479}{22,4}\approx0,11\left(mol\right)\)

PTHH: \(SO_2+Ba\left(OH\right)_2--->BaSO_3\downarrow+H_2O\)

Theo PT: \(n_{BaSO_3}=n_{SO_2}=0,11\left(mol\right)\)

=> \(m_{BaSO_3}=0,11.217=23,87\left(g\right)\)

Câu 4 : 

\(n_{SO2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Pt : \(SO_2+Ba\left(OH\right)_2\rightarrow BaSO_3+H_2O|\) 

        1              1                  1            1

     0,15          0,15             0,15

a) \(n_{Ba\left(OH\right)2}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)

⇒ \(m_{Ba\left(OH\right)2}=0,15.171=25,65\left(g\right)\)

\(C_{ddBa\left(OH\right)2}=\dfrac{25,65.100}{150}=17,1\)⁰/₀

b) \(n_{BaSO3}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)

⇒ \(m_{BaSO3}=0,15.217=32,55\left(g\right)\)

c) Pt : \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O|\)

                  1               2            1              2

                0,15          0,3

\(n_{HCl}=\dfrac{0,15.2}{1}=0,3\left(mol\right)\)

\(m_{HCl}=0,3.36,5=10,95\left(g\right)\)

\(m_{ddHCl}=\dfrac{10,95.100}{20}=54,75\left(g\right)\)

\(V_{ddHCl}=\dfrac{54,75}{1,2}=45,625\left(ml\right)\)

 Chúc bạn học tốt

\(a)Ba\left(OH\right)_2+Na_2CO_3\rightarrow2NaOH+BaCO_3\\ n_{Ba\left(OH\right)_2}=0,2.2=0,4mol\\ n_{BaCO_3}=n_{Na_2CO_3}=n_{Ba\left(OH\right)_2}=0,4mol\\ m_{BaCO_3}=0,4.171=68,4g\\ b)V_{Na_2CO_3}=\dfrac{0,4}{1}=0,4l\\ c)n_{NaOH}=2n_{Ba\left(OH\right)_2}=0,8mol\\ C_{M\left(NaOH\right)}=\dfrac{0,8}{0,2+0,4}=\dfrac{4}{3}M\)

\(n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: Ba(OH)₂ + SO₂ --> BaSO₃ + H₂O

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

=> m_BaSO3 = 0,3.217 = 65,1 (g)

`n_(SO_2) = (6.72)/(22.4) = 0,3 mol`.

Ta có: `Ba(OH)_2 + SO_2 -> BaSO_3 + H_20`

`=> n_(BaSO_3) = 0,3 xx 1 : 1 = 0,3`.

`=> m_(BaSO_3) = 0,3 xx 217 = 65,1 g`.

1:

a) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

______0,2------>0,2------------------->0,2_____(mol)

=> \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b) \(V_{ddH_2SO_4}=\dfrac{0,2}{1}=0,2\left(l\right)\)

2:

a) 

\(n_{HCl}=2.0,2=0,4\left(mol\right)\)

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

______0,2<------0,4------------------>0,2______(mol)

=> \(m_{Mg}=0,2.24=4,8\left(g\right)\)

b) \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)