\(n_{SO2}=\frac{0,15}{n_{OH}}=0,5.\left(0,2+0,2\right)\Rightarrow\) Tạo 2 muối

\(n_{SO2/3}=\frac{a}{n_{HSO3}}=b\)

\(\left\{{}\begin{matrix}a+b=0,15\\2a=b=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,05\\b=0,1\end{matrix}\right.\)

m_muối = m_Na⁺ + m_K⁺ + m_SO2/3 + m_HSO3

\(\Rightarrow m_{muoi}=0,1.23+0,1.39.80+0,1.81=18,3\left(g\right)\)

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

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

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

=> 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 : 

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

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

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

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

Đáp án A

Bài 7 :

\(n_{CO_2}=\frac{V}{22,4}=\frac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH sẽ xảy ra :

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

\(NaOH+CO_2\rightarrow NaHCO_3\) ( II )

a, \(n_{NaOH}=C_M.V=1.0,3=0,3\left(mol\right)\)

=> \(T=\frac{0,3}{0,1}=3>2\)

=> Xảy ra phản ứng ( I ) NaOH dư .

=> \(m_{ct}=m_{Na_2CO_3}+m_{NaOH}\)_dư

= \(0,1.106+40\left(0,3-0,2\right)=14,6\left(g\right)\)

b, \(n_{NaOH}=C_M.V=1.0,125=0,125\left(mol\right)\)

=> \(T=1< \frac{0,125}{0,1}=1,25< 2\)

=> Xảy ra phản ứng ( I ) ( II ) .

\(PTHH:5NaOH+4CO_2\rightarrow3NaHCO_3+Na_2CO_3+H_2O\)

.................0,125............0,1..........0,075..............0,025...................

=> \(m_{CT}=m_{NaHCO_3}+m_{Na_2CO_3}=8,375\left(g\right)\)

c, \(n_{NaOH}=C_M.V=1,5.0,05=0,075\left(mol\right)\)

=> \(T=\frac{0,075}{0,1}=0,75< 1\)

=> Xảy ra phản ứng ( II ), CO₂ dư

=> \(m_{ct}=m_{NaHCO_3}=6,3\left(g\right)\)