\(n_{H_2S}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: 2NaOH + H₂S --> Na₂S + 2H₂O

                             0,1---->0,1

=> m_Na2S = 0,1.78 = 7,8 (g)

\(n_{CO_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

\(n_{NaOH}=\dfrac{8}{40}=0.2\left(mol\right)\)

\(T=\dfrac{0.2}{0.2}=1\)

\(\Rightarrow\text{Tạo muối axit}\)

\(NaOH+CO_2\rightarrow NaHCO_3\)

\(0.2...............0.2............0.2\)

\(m_{NaHCO_3}=0.2\cdot84=16.8\left(g\right)\)

Ta có: \(n_{CO_2}=0,2\left(mol\right)\)

\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

\(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}=1\) ⇒ Pư tạo muối NaHCO₃.

PT: \(CO_2+NaOH\rightarrow NaHCO_3\)

___0,2________________0,2 (mol)

⇒ m_NaHCO3 = 0,2.84 = 16,8 (g)

Bạn tham khảo nhé!

\(\dfrac{nNaOH}{nCO2}=\dfrac{\dfrac{8}{40}}{\dfrac{4,48}{22,4}}=1\)=>tạo muối NaHCO3

pthh CO2+NaOH->NaHCO3=>nNaHCO3=NCO2=0,2mol

=>mNaHCO3=0,2.84=16,8 g

Na2CO3 + H2SO4 -> Na2SO4 + CO2 + H2O

CO2 + 2NaOH -> Na2CO3 + H2O 

a             2a           a

CO2 + NaOH -> NaHCO3

b            b                 b

nNaOH= \(\frac{8\cdot40\%}{40}\)= 0,08 mol

Theo đề bài ta có hệ pt

2a+b=0,08

106a+84b= 5,48

=> a=0,02        b=0,04

%mNa2CO3 = \(\frac{0,02\cdot106}{5,48}\)*100% = 38,69%

%mNaHCO3 = 61,31%

nCO2 = 0,02+0,04=0,06

CMH2SO4 = 0,06/0,2=0,3 M

a) \(n_{SO_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

\(n_{NaOH}=0,5.2=1\left(mol\right)\)

PTHH: 2NaOH + SO₂ --> Na₂SO₃ + H₂O

               1------->0,5------>0,5

             Na₂SO₃ + SO₂ + H₂O --> 2NaHSO₃

                0,1<-----0,1--------------->0,2

=> Thu được muối Na₂SO₃, NaHSO₃

\(\left\{{}\begin{matrix}m_{Na_2SO_3}=0,4.126=50,4\left(g\right)\\n_{NaHSO_3}=0,2.104=20,8\left(g\right)\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}C_{M\left(Na_2SO_3\right)}=\dfrac{0,4}{0,5}=0,8M\\C_{M\left(NaHSO_3\right)}=\dfrac{0,2}{0,5}=0,4M\end{matrix}\right.\)

\(n_{H_2S}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(n_{NaOH}=0.3\cdot0.5=0.15\left(mol\right)\)

\(\dfrac{n_{NaOH}}{n_{H_2S}}=\dfrac{0.15}{0.1}=1.5\)

\(\rightarrow\text{Phản ứng tạo ra 2 muối}\)

\(n_{Na_2S}=a\left(mol\right),n_{NaHS}=b\left(mol\right)\)

\(\left\{{}\begin{matrix}2a+b=0.15\\a+b=0.1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=0.05\\b=0.05\end{matrix}\right.\)

\(m_{\text{Muối}}=0.05\cdot78+0.05\cdot56=6.7\left(g\right)\)

\(n_{H_2S}=0,08\left(mol\right),n_{OH^-}=0,1.0,05+0,1.0,08=0,013\left(mol\right)\)

T=\(\dfrac{0,013}{0,08}=0,1625\)=> Tạo 1 muối HS- , H2S dư

Muối gồm KHS và NaHS

=> \(m_{muối}=0,1.0,05.56+0,1.0,08.72=0,856\left(g\right)\)

a) \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: CO₂ + 2NaOH → Na₂CO₃ + H₂O

Mol:      0,15       0,3              0,15

\(C_{M_{ddNaOH}}=\dfrac{0,3}{0,2}=1,5M\)

b) Na₂CO₃: natri cacbonat

\(m_{Na_2CO_3}=0,15.106=15,9\left(g\right)\)

c) 

PTHH: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Mol:       0,15          0,075

\(V_{ddH_2SO_4}=\dfrac{0,075}{1}=0,075\left(l\right)=75\left(ml\right)\)

\(n_{CO2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

a) Pt : \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O|\)

             1            2                   1               1

           0,15       0,3                0,15

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

200ml = 0,2l

\(C_{M_{ddNaOH}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

b \(n_{Na2CO3}=\dfrac{0,3.1}{2}=0,15\left(mol\right)\)

⇒ \(m_{Na2CO3}=0,15.106=15,9\left(g\right)\)

Ten muoi thu duoc : natri cacbonat 

c) Pt : \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O|\)

                 2               1               1                2

              0,3              0,15

\(n_{H2SO4}=\dfrac{0,3.1}{2}=0,15\left(mol\right)\)

\(V_{ddH2SO4}=\dfrac{0,15}{1}=0,15\left(l\right)=150\left(ml\right)\)

 Chuc ban hoc tot