Câu 4 :
\(MnO_4+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)
Ta có :
\(n_{MnO2}=\frac{139,2}{87}=1,6\left(mol\right)\)
\(\Rightarrow n_{Cl2}=n_{MnO2}=1,6\left(mol\right)\)
\(\Rightarrow V_{Cl2}=1,6.22,4=35,84\left(l\right)\)
Câu 4 :
\(Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O\)
\(n_{Cl2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(\Rightarrow n_{NaOH}=2n_{Cl2}=0,3.2=0,6\left(mol\right)\)
\(\Rightarrow V_{NaOH}=\frac{0,6}{1}=0,6\left(l\right)=600\left(ml\right)\)