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
