1. 2Al+3Cl2\(\rightarrow\)2AlCl3
nAlCl3=\(\frac{26,7}{133,5}\)=0,2
\(\rightarrow\)nAl=0,2\(\rightarrow\)mAl=0,2.27=5,4
\(\rightarrow\)nCl2=0,3\(\rightarrow\)VCl2=0,3.22,4=6,72l
2. 2KMnO4+16HCl\(\rightarrow\)2KCl+2MnCl2+5Cl2+8H2O
nFeCl3=\(\frac{16,25}{162,5}\)=0,1
2Fe+3Cl2\(\rightarrow\)2FeCl3
\(\rightarrow\)nCl2=0,1.1,5=0,15
\(\rightarrow\)nKMnO4=\(\frac{0,15.2}{5}\)=0,06
\(\rightarrow\)mKMnO4=0,06.158=9,48
\(\rightarrow\)nHCl=0,48
\(\rightarrow\)VddHCl=0,48l=480ml