Question

A hydrogen-filled balloon was ignited and 1.40 g of hydrogen reacted with 11.2 g of oxygen

Answer 1

2H₂  +  O₂  ------->  2H₂O

mole of H₂  =   $\frac{mass of H_{2} }{molar mass of H_{2}}$
   
                   =  $\frac{1.40 g}{(2 * 1 g / mol}$
 
                   =  0.7 mol

mole ratio of H₂  :  H₂O
                  =  1   :    1

∴ if mole  of H₂O  = 0.7 mol
 then mole of water  =  0.7 mol

Thus:
  mass of water  =  mole x  Mr
                           =  0.7 mol  x  ((2 x 1) + (1 x 16))
                           =  12.6 g of water vapour