In the explosion of a hydrogen-filled balloon, 0.80 g of hydrogen reacted with 6.4 g of oxygen. How many grams of water vapor are formed? (Water vapor is the only product.)

0.4 moles H2O x (18.0 g / mole) = 7.2 g H2O

Explanation:

write the values given in the question

In an explosion of hydrogen balloon, 0.80g of hydrogen is reacted with 6.4g of oxygen.

Write the balanced equation

2 H2 + O2 - --> 2 H2O

0.80 g H2 x (1 mole / 2g) = 0.4 moles H2

6.4 g O2 x (1 mole / 32g) = 0.2 moles O2

from balanced equation, 2 moles H2 react with 1 mole O2

therefore

0.4 moles H2 x (1 mole O2 / 2 moles H2) = 0.2 moles O2

since we need 0.2 moles O2 to react with all the H2 and we have exactly that amount, the amounts are said to be

"stoichiometric" and either reactant can be considered limiting

from the balanced equation, 2 moles H2 - -> 2 moles H2O ... therefore

0.4 moles H2 x (2 moles H2O / 2 moles H2) = 0.4 moles H2O

Then

0.4 moles H2O x (18.0 g / mole) = 7.2 g H2O