Three moles of $B_2{H}_6$  are completely reacted with methanol, the number of moles of boron containing product formed is _____

 $B_2{H}_6+6C{H}_3-OH\to 2B{\left(OC{H}_3\right)}_3+6H_2$
1 Mole of  $B_2{H}_6$ gives 2 moles of  $B{\left(OC{H}_3\right)}_3$  . So, 3 moles of $B_2{H}_6$   would gives 6 moles of  $B{\left(OC{H}_3\right)}_3$ .