NOTE:- i have taken arbitrary atomic mass to make it easier to explain.learn about limiting reagent and excess reagent once you finish reading this. hope this helps you.
let atomic mass of boron be = 2g/mol
hydrogen = 1g/mol
oxygen = 3g/mol
therefore, molecular mass of B5H9= 19g // i will takr it to be 20, for easy explanation
O2 = 6g
H2O = 5g
now,
no. of moles of B5H9 taken = 126/20=6.3
no. of moles of O2 taken = 192/6= 32
according to he rxn.
2 moles of B5H9 will require 12 moles of O2
so 6.3 " " " " " (6*6.3)" " " =37.8 =38(approx)
but we have only 32 moles of O2. hence O2 is the limiting reagent.32 moles of O2 will completly react.
gong back to the rxn.
12 moles of O2 will give 9 moles of water
32 " " " " " 24 " " "
therefore amuont of water produced= 24*5=120 g.