\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2mol\\ n_{KOH}=0,3.1=0,3mol\\ T=\dfrac{0,3}{0,2}=1,5\)

⇒1<T<2, tạo muối KHCO₃, K₂CO₃

\(n_{K_2CO_3}=a;n_{KHCO_3}=b\\ 2KOH+CO_2\rightarrow K_2CO_3+H_2O\\ KOH+CO_2\rightarrow KHCO_3\\ \Rightarrow\left\{{}\begin{matrix}2a+b=0,3\\a+b=0,2\end{matrix}\right.\\ \Rightarrow a=b=0,1\\ m_{muối}=0,1\left(138+100\right)=23,8g\)

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

=> 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: \(\left\{{}\begin{matrix}a+b=0,2\\2a+b=0,25\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,15\end{matrix}\right.\)

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

a, \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

b, \(n_{H_2SO_4}=0,2.1,5=0,3\left(mol\right)\)

Theo PT: n_Na2SO4 = n_H2SO4 = 0,3 (mol) ⇒ m_Na2SO4 = 0,3.142 = 42,6 (g)

n_NaOH = 2n_H2SO4 = 0,6 (mol) ⇒ m_NaOH = 0,6.40 = 24 (g)

c, \(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

\(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,6}{0,4}=1,5\) → Pư tạo NaHCO₃ và Na₂CO₃.

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

\(NaOH+CO_2\rightarrow NaHCO_3\)

Đáp án D

Số mol các chất là

Bài 1 : 

\(n_{CO2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Pt : \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O|\)

          1              1                 1            1

        0,2           0,2               0,2

a) \(n_{CaCO3}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{CaCO3}=0,2.100=20\left(g\right)\)

b) \(n_{Ca\left(OH\right)2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

200ml = 0,2l

\(C_{M_{ddCa\left(OH\right)2}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

 Chúc bạn học tốt

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