\(\text{a, }n_{K_2SO_3}=\dfrac{m}{M}=\dfrac{39,5}{158}=0,25\left(mol\right)\)
\(n_{HCl}=\dfrac{m}{M}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
\(\text{PTHH : }K_2SO_3+2HCL\rightarrow2KCl+H_2O+SO_2\)
\(\text{Trước pư : 0,25}\) \(\text{0,4}\)
\(\text{Trong pư : }\dfrac{0,25}{1}\) \(>\) \(\dfrac{0,4}{2}\)
\(\text{Sau pư : }\) \(0,1\) \(\text{0,2}\) \(0,1\)
\(V_{SO_2}=22,4.n=22,4.0,1=2,24\left(l\right)\)
\(b,m_{HCl}\text{pư}=n.M=0,2.36,5=7,3\left(g\right)\)
\(\Rightarrow m_{HCl}\text{dư}=14,6-7,3=7,3\left(g\right)\)
\(n_{K_2SO_3}=\dfrac{39,5}{158}=0,25\left(mol\right)\)
\(n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
PTHH: \(K_2SO_3+2HCl\rightarrow2KCl+H_2O+SO_2\)
Theo PTHH ta có: \(\dfrac{0,25}{1}>\dfrac{0,4}{2}=0,2\)
\(\Rightarrow K_2SO_3\) dư, HCl hết. Vậy ta tính theo \(n_{HCl}\)
Theo PT: \(n_{SO_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\)
a. \(\Rightarrow V_{SO_2}=0,2.22,4=4,48\left(l\right)\)
b. Theo PT ta có:
\(n_{K_2SO_3\left(pư\right)}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)
\(\Rightarrow n_{K_2SO_3\left(dư\right)}=0,25-0,2=0,05\left(mol\right)\)
\(\Rightarrow m_{K_2SO_3}=0,05.158=7,9\left(g\right)\)