Hydroboration oxidation of propene gives 1-propanol as the product. Number of moles of alcohol formed when one mole of BH₃ / THF is involved in the reaction is

$$\begin{gathered} {\mathrm{CH}}_3-\mathrm{CH}={\mathrm{CH}}_2\underset{\mathrm{THF}}{\overset{{\mathrm{BH}}_3}{\to }}{\mathrm{CH}}_3-\underset{\stackrel{|}{\mathrm{H}}}{\mathrm{C}}\mathrm{H}-\underset{\stackrel{|}{{\mathrm{BH}}_2}}{\mathrm{C}}{\mathrm{H}}_2 \\ \left({\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\right){\mathrm{BH}}_2\stackrel{2 {\mathrm{CH}}_3\mathrm{CH}={\mathrm{CH}}_2}{\to }\underset{\mathrm{trialkyl} \mathrm{borane}}{{\left({\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\right)}_3\mathrm{B}} \\ {\left({\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\right)}_3\mathrm{B}\underset{{\mathrm{OH}}^{-}}{\overset{{\mathrm{H}}_2{\mathrm{O}}_2}{\to }}3{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{OH} \end{gathered}$$

Thus 1 mole BH₃ require 3 moles of alkene and 3 moles of H₂O₂ to give 3 moles of alcohol.