Question

How many moles of water are produced from 19.2 g of B2H6

Answer 1

The Balanced chemical equation of reaction of Borane with oxygen is as follow,
                                  B₂H₆ + 3O₂ -----> 2HBO₂ + 2H₂O
According to this equation 27.66 g (1 mole) of B₂H₆ reacts with oxygen to produce 36 g (2 moles) of water.
The amount of water produced when 19.2 g of B₂H₆ reacted is calculated as follow,
$\frac{27.66 g B2H6 produced}{19.2 g of B2H6 will produce}$ = $\frac{36 g of H2O}{x g of water}$
Solving for x,                       
              x = (36 g of H₂O ₓ 19.2 g of H₂B₆) / 27.66 g of B₂H₆

              x = 24.98 g of H₂O

Result:
           24.98 g  of water is produced when 19.2 g of B₂H₆ is reacted with excess of oxygen.