\(n_{SO_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

\(n_{OH^-}=0.5\cdot0.2+0.5\cdot0.2=0.2\left(mol\right)\)

\(T=\dfrac{0.2}{0.15}=1.33\)

=> Tạo ra \(SO_3^{2-},HSO_3^-\)

Đặt :

\(n_{SO_3^{2-}}=a\left(mol\right),n_{HSO_3^{2-}}=b\left(mol\right)\)

Ta có hệ phương trình : 

\(\left\{{}\begin{matrix}n_S=a+b=0.15\left(mol\right)\\n_{OH^-}=2a+b=0.2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=0.05\\b=0.1\end{matrix}\right.\)

\(m_{Muối}=m_{Na^+}+m_{K^+}+m_{SO_3^{2-}}+m_{HSO_3^-}\)

\(=0.5\cdot0.2\cdot23+0.5\cdot0.2\cdot39+0.05\cdot80+0.1\cdot81=18.3\left(g\right)\)

\(n_{CO_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

\(n_{NaOH}=\dfrac{12}{40}=0.3\left(mol\right)\)

\(T=\dfrac{0.3}{0.15}=2\)

\(\Rightarrow\text{Tạo ra muối trung hòa}\)

\(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)

\(0.3...........0.15..............0.15\)

\(m_{Na_2CO_3}=0.15\cdot106=15.9\left(g\right)\)

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

\(n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\)

\(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}=2\). Vậy: Pư tạo muối trung hòa Na₂CO₃.

PT: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

___0,15______________0,15 (mol)

⇒ m_Na2CO3 = 0,15.106 = 15,9 (g)

Bạn tham khảo nhé!

tỉ lệ mol: \(\dfrac{nNaOH}{nCO2}=\dfrac{\dfrac{12}{40}}{\dfrac{3,36}{22,4}}=2\)=>tạo muối Na2CO3

pthh: CO2+2NaOH->Na2CO3+H2O

=>nNa2CO3=nCO2=0,15mol=>mNa2CO3=0,15.106=15,9 gam

vậy Tính khối lượng muối tan trong dd X là 15,9 g

nSO2 = 3.36/22.4 = 0.15 (mol) 

nNaOH = 0.5 * 0.1 = 0.05 (mol) 

nNaOH / nSO2 = 0.05 / 0.15 = 0.3 

Chỉ tạo ra muối axit

NaOH + SO2 => NaHSO3

0.05.........0.05.........0.05

mNaHSO3 = 0.05 * 104 = 5.2 (g) 

nSO2= 3,36/22,4=0,15mol

nNaOH=0,5*0,1=0,05 mol

pthh   2NaOH + SO2 --> Na2SO3 + H2O

              0,05                       0,025                         mol

ta thấy nNaOH/2<nSO2/1 =>NaOH hết , SO2 dư

=> m Na2SO3=0,025*126=3,15 g

Bài 7:

Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\n_{NaOH}=0,2\cdot1=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo 2 muối

PTHH: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

                 a_______2a__________a              (mol)

            \(CO_2+NaOH\rightarrow NaHCO_3\)

                 b_______b__________b       (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}a+b=0,15\\2a+b=0,2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,1\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{CO_2}+m_{ddNaOH}=0,15\cdot44+200\cdot1,25=256,6\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Na_2CO_3}=\dfrac{0,05\cdot106}{256,6}\cdot100\%\approx2,1\%\\C\%_{NaHCO_3}=\dfrac{0,1\cdot72}{256,6}\cdot100\%\approx2,8\%\end{matrix}\right.\)

Bài 8:

PTHH: \(RCO_3+2HNO_3\rightarrow R\left(NO_3\right)_2+CO_2\uparrow+H_2O\)

Giả sử \(n_{RCO_3}=1\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{HNO_3}=2\left(mol\right)\\n_{R\left(NO_3\right)_2}=1\left(mol\right)=n_{CO_2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ddHNO_3}=\dfrac{2\cdot63}{20\%}=630\left(g\right)\\m_{R\left(NO_3\right)_2}=R+124\left(g\right)\\m_{CO_2}=44\left(g\right)\end{matrix}\right.\) \(\Rightarrow C\%_{R\left(NO_3\right)_2}=\dfrac{124+R}{R+60+630-44}=0,26582\)

\(\Leftrightarrow R=65\) (Kẽm) \(\Rightarrow\) CTHH của muối cacbonat là ZnCO₃

nCO2=0,15mol

nNa2CO3=0,25mol

PTHH: CO2+2NaOH=>Na2CO3 + H2O

            0,15:                       0,25    =>nNa2CO3 dư

p/ư:        0,15----0,3---------->0,15

=>V=0,3:1,2=0.25lit

=>  CM(Na2CO3)=0,15:0,25=0,6M