Question

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

How many grams of water vapor were formed? (Assume that water vapor is the only product.)

Answer 1

2H₂ + O₂ -------> 2H₂O

mole of H₂ =
`(mass of H_(2) )/(molar mass of H_(2))`

=
`(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