PTHH: \(K_2SO_3+2HCl\rightarrow2KCl+H_2O+SO_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{K_2SO_3}=\dfrac{39,5}{158}=0,25\left(mol\right)\\n_{HCl}=\dfrac{200\cdot10,95\%}{36,5}=0,6\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,25}{1}< \dfrac{0,6}{2}\) \(\Rightarrow\) HCl còn dư, tính theo K2SO3
\(\Rightarrow\left\{{}\begin{matrix}n_{SO_2}=0,25\left(mol\right)\\n_{KCl}=0,5\left(mol\right)\\n_{HCl\left(dư\right)}=0,6-0,25\cdot2=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{SO_2}=0,25\cdot22,4=5,6\left(l\right)\\m_{SO_2}=0,25\cdot64=16\left(g\right)\\m_{KCl}=0,5\cdot74,5=37,25\left(g\right)\\m_{HCl\left(dư\right)}=0,1\cdot36,5=3,65\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd\left(sau.pư\right)}=m_{K_2SO_3}+m_{ddHCl}-m_{SO_2}=223,5\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{KCl}=....\\C\%_{HCl\left(dư\right)}=....\end{matrix}\right.\)
