a, \(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)
Ta có :
\(n_{MnO2}=\frac{34,8}{87}=0,4\left(mol\right)\)
\(n_{HCl}=\frac{200.36,5\%}{36,6}=2\left(mol\right)\)
Tỉ lệ : \(\frac{0,4}{1}< \frac{2}{4}\)
\(\Rightarrow\) HCl dư,tính theo MnO2
\(n_{Cl2}=n_{MnO2}=0,4\left(mol\right)\)
\(H=80\%\)
\(V_{Cl2}=0,4.80\%.22,4=7,168\left(l\right)\)
b, \(n_{NaOH}=0,5.2=1\left(mol\right)\)
\(Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O\)
\(n_{NaOH}=1-0,32.2=0,36\left(mol\right)\)
\(CM_{NaOH}=\frac{0,36}{0,5}=0,72M\)
\(n_{NaCl}=n_{NaClO}=0,32\left(mol\right)\)
\(\Rightarrow CM_{NaCl}=CM_{NaClO}=\frac{0,32}{0,5}=0,64M\)
