Three moles of $\mathrm{B}_{2} \mathrm{H}_{6}$ are completely reacted with methanol. The number of m

Question

Three moles of $\mathrm{B}_{2} \mathrm{H}_{6}$ are completely reacted with methanol. The number of moles of boron containing product formed is $\qquad$ -\

Professor PC Thomas Classes · Answer

(6) : $\left[\mathrm{B}_{2} \mathrm{H}_{6}+6 \mathrm{CH}_{3} \mathrm{OH} \longrightarrowight.$ $\left.2 \mathrm{~B}\left(\mathrm{OCH}_{3}ight)_{3}+6 \mathrm{H}_{2}ight] 	imes 3$\ $3 \mathrm{~B}_{2} \mathrm{H}_{6}+18 \mathrm{CH}_{3} \mathrm{OH} \longrightarrow 6 \mathrm{~B}\left(\mathrm{OCH}_{3}ight)_{3}$ +18 $\mathrm{H}_{2}$  Thus, no. of moles of boron containing product formed by reacting 3 moles of $\mathrm{B}_{2} \mathrm{H}_{6}$ completely with methanol is 6 .\