a) \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(n_{NaOH}=0,3.1=0,3\left(mol\right)\)

Xét tỉ lệ \(\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,3}{0,2}=1,5\)

=> Tạo ra muối NaHCO₃, Na₂CO₃

b) 

PTHH: 2NaOH + CO₂ --> Na₂CO₃ + H₂O

             0,3----->0,15------>0,15

            Na₂CO₃ + CO₂ + H₂O --> 2NaHCO₃

             0,05<---0,05---------------->0,1

=> \(\left\{{}\begin{matrix}m_{Na_2CO_3}=\left(0,15-0,05\right).106=10,6\left(g\right)\\n_{NaHCO_3}=0,1.84=8,4\left(g\right)\end{matrix}\right.\)

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(n_{Ca\left(OH\right)_2}=0,1.0,5=0,05\left(mol\right)\)

Xét \(\dfrac{n_{CO_2}}{n_{Ca\left(OH\right)_2}}=\dfrac{0,2}{0,05}=4\)

=> Tạo ra muối HCO₃^-

PTHH: Ca(OH)₂ + 2CO₂ --> Ca(HCO₃)₂

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

=> m_Ca(HCO3)2 = 0,05.162=8,1(g)

\(n_{NaOH}=0,3.1=0,3\left(mol\right)\\ n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\1< \dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,3}{0,2}=1,5< 2\)

=> Sản phẩm tạo 2 muối: Na₂CO₃ và NaHCO₃

=> CHỌN A

A

chọn A

Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) 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: \(\left\{{}\begin{matrix}a+b=0,2\\2a+b=0,25\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,15\end{matrix}\right.\)

\(\Rightarrow m_{muối}=m_{Na_2CO_3}+m_{NaHCO_3}=0,05\cdot106+0,15\cdot84=17,9\left(g\right)\)

nCO2=0,3(mol)

nNaOH=0,5(mol)

Vì: 1< nNaOH/nCO2=0,5/0,3=1,67<2

=> Sp thu được là hỗn hợp 2 muối

PTHH: 2 NaOH + CO2 -to-> Na2CO3 + H2O

2x____________x_________x(mol)

NaOH + CO2 -> NaHCO3

y_____y______y(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}2x+y=0,5\\x+y=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

=> m(muối)= mNa2CO3 + mNaHCO3=0,2.106+0,1.84= 29,6(g)

\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

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

Mol:            0,3                             0,3

\(\Rightarrow m_{Na_2CO_3}=0,3.106=31,8\left(g\right)\)

n_CO2= \(\dfrac{V}{22,4}\)=\(\dfrac{6,72}{22,4}\)=0,3 (mol).

n_NaOH= C_M.V= 1 . 0,5= 0,5(mol).

\(\dfrac{n_{NaOH}}{n_{CO2}}\)=\(\dfrac{0,5}{0,3}\)= 1,67 mà 1<1,67<2 ⇒ tạo 2 muối (1) và (2).

CO₂ + NaOH -> NaHCO₃   (1)

CO₂ + NaOH -> Na₂CO₃ + H₂O  (2)

Lập TLSS: \(\dfrac{0,3}{1}\)  >  \(\dfrac{0,5}{0,1}\)

⇒Tính theo NaOH.

m_NaHCO3= 0,5 . 84 = 42g

m_Na2CO3= 0,5 . 106= 53g

a, \(Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+SO_2+H_2O\)

\(K_2SO_3+H_2SO_4\rightarrow K_2SO_4+SO_2+H_2O\)

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

Theo PT: \(n_{H_2SO_4}=n_{SO_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,3.98}{20\%}=147\left(g\right)\)

b, Ta có: 126n_Na2SO3 + 158n_K2SO3 = 44,2 (1)

Theo PT: \(n_{SO_2}=n_{Na_2SO_3}+n_{K_2SO_3}=0,3\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Na_2SO_3}=0,1\left(mol\right)\\n_{K_2SO_3}=0,2\left(mol\right)\end{matrix}\right.\)

Có: m _{dd sau pư} = 44,2 + 147 - 0,3.64 = 172 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Na_2SO_3}=\dfrac{0,1.126}{172}.100\%\approx7,33\%\\C\%_{K_2SO_3}=\dfrac{0,2.158}{172}.100\%\approx18,37\%\end{matrix}\right.\)

c, \(n_{Ba\left(OH\right)_2}=0,5.1=0,5\left(mol\right)\)

\(\Rightarrow\dfrac{n_{SO_2}}{n_{Ba\left(OH\right)_2}}=0,6< 1\) → Pư tạo BaSO₃.

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

\(n_{BaSO_3}=n_{SO_2}=0,3\left(mol\right)\Rightarrow m_{BaSO_3}=0,3.217=65,1\left(g\right)\)