\(V_{H_2}=0,8.22,4=17,92\left(l\right)\)
\(n_{Cl_2}=\dfrac{2,479}{22,4}\approx0,1\left(mol\right)\)
\(V_{H_2}=0,8.22,4=17,92\left(l\right)\\ n_{Cl_2}=\dfrac{2,479}{22,4}=0,111\left(mol\right)\)
\(-V_{H_2}=0,8.22,4=17,92\left(l\right)\)
\(-n_{Cl_2}=\dfrac{2,479}{22,4}\approx0,11\left(mol\right)\)