a, PT: \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)
\(K_2SO_3+2HCl\rightarrow2KCl+SO_2+H_2O\)
b, Ta có: \(n_{CO_2}+n_{SO_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{CO_2}+2n_{SO_2}=0,03.2=0,06\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,06}{0,5}=0,12\left(M\right)\)
c, Gọi: \(\left\{{}\begin{matrix}n_{Na_2CO_3}=x\left(mol\right)\\n_{K_2SO_3}=y\left(mol\right)\end{matrix}\right.\)
⇒ 106x + 158y = 4,22 (1)
Theo PT: \(n_{CO_2}+n_{SO_2}=n_{Na_2CO_3}+n_{K_2SO_3}=x+y=0,03\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,01\left(mol\right)\\y=0,02\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na_2CO_3}=\dfrac{0,01.106}{4,22}.100\%\approx25,12\%\\\%m_{K_2SO_3}\approx74,88\%\end{matrix}\right.\)
