A sample of H2 with a volume of 35.0 L at STP is reacted with excess O2 to produce H2O. How many grams of H2O are produced?
2 H2 (g) + O2 (g) --> 2 H2O (g)
This is a stoichiometry problem. Here is a worked example. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
However, when all are gases, one may take a shortcut and use L as if they were moles. Try it, you'll like it.
LH2O = LH2 x (2 moles H2O/2 moles H2) = 35L H2 x (2/2) = 35 L H2O
Then moles H2O = 35/22.4 = ? moles and
grams = moles x molar mass.
To determine the mass of H2O produced, we need to use stoichiometry. The balanced chemical equation provides the mole-to-mole ratios between the reactants and products:
2 H2 (g) + O2 (g) → 2 H2O (g)
From the equation, we can see that 2 moles of H2 will produce 2 moles of H2O.
First, we need to convert the given volume of H2 at STP (Standard Temperature and Pressure) to moles. At STP, 1 mole of any gas occupies a volume of 22.4 liters.
35.0 L H2 x (1 mole H2 / 22.4 L H2) = 1.56 moles H2
Since 2 moles of H2 will produce 2 moles of H2O, we can deduce that 1.56 moles of H2 will produce the same amount of H2O.
To calculate the molar mass of H2O, we need to know the atomic masses: H2 = 2.02 g/mol, O = 16.00 g/mol.
H2O = (2 x H) + (1 x O) = (2 x 1.01 g/mol) + (1 x 16.00 g/mol) = 18.02 g/mol
Finally, we can calculate the mass of H2O produced:
Mass of H2O = moles of H2O x molar mass of H2O
= 1.56 moles H2O x 18.02 g/mol
= 28.11 g H2O
Therefore, 28.11 grams of H2O are produced.