a
\((a)\) In order to maximize the amount of monohalogenated product obtained,a radical substitution reaction should be carried out in the presence of excess of alkane. The presence of excess alkane in the reaction mixture ensures that  there is a greater probability of the halogen radical colliding with a molecule of  alkane than with a molecule of alkyl halide. This is true even toward . the end  of the reaction, by which time a considerable amount of alkyl halide will have been formed. If the halogen radical abstracts a hydrogen from a molecule of  alkyl halide rather than from a molecule of alkane, a dihalogenated product will  be obtained 

\(C{l^ \bullet } + C{H_3}Cl \to {\,^ \bullet }C{H_2}Cl + HCl\)

\(^ \bullet C{H_2}Cl + C^{12} \to C{H_2}C{l_2} + C{l^ \bullet }\)

Bromination of alkanes follows the same mechanism as chlorination. The only  difference is that chlorination produces alkyl chlorides, whereas bromination  forms alkyl bromides