a, \(n_{K_2SO_3}=\dfrac{39,5}{158}=0,25\left(mol\right);n_{HCl}=0,2.2=0,4\left(mol\right)\)
PTHH: K2SO3 + 2HCl → 2KCl + SO2 + H2O
Mol: 0,2 0,4 0,4 0,2
b, Ta có: \(\dfrac{0,25}{1}>\dfrac{0,4}{2}\) ⇒ K2SO3 dư, HCl hết
\(m_{K_2SO_3}=\left(0,25-0,2\right).158=7,9\left(g\right)\)
c, \(V_{SO_2}=0,2.22,4=4,48\left(l\right)\)
d, Sau pứ dd chứa K2SO3 , KCl
\(m_{KCl}=0,4.74,5=29,8\left(g\right)\)