Hello, Denim,
We need to start with a balanced equation for this (explosive) reaction. I dislike wasting hydrogen and good oxygen just to make water, but it provides a nice bang for the buck, and water is becoming increasingly valuable. Here's a balanced equation:
2H2 + 1O2 = 2H2O
It's balanced since we have 4H atoms and 2 O atoms going in, and the same number going out in the products.
We can read this equation by stating 2 molecules of hydrogen react with 1 molecule of oxygen to make 2 molecules of water. Since the mole is a counting unit, we can also say 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water. If we use the mole unit, it allows us to translate the actual numbers of grams of each. The molar mass of an atom or molecule is the mass, in grams, that it takes to have 6.02x1023 particles of that atom/molecule. Like all lazy chemists, I prefer to simply write 1 and 2 moles. Not 6.02x1023 and 1.204x1024 molecules.
In this case we are given 10.5 of hydrogen and 36.9 grams of oxygen. Convert those bvalues into moles:
H2: (10.5 g)/(2 g/mole) = 5.25 moles
O2: (36.9 g)/(32 g/mole) = 1.153 moles
Note that although we have over 4 times the mass of oxygen than of hydrogen, we only have about 1/5 the number of moles. Granted, the equation says we need 2 hydrogens for every 1 oxygen, but it is still clear that we won't have enough oxygen molecules to react with all the hydrogen molecules.
Oxygen thus becomes the "limiting reagent." The reaction will stop once the O2 molecules are consumed. We need to base our prediction of water on the amount of O2 we are given (1.153 moles).
The equation states we should expect 2 moles of H2O for every 1 mole of O2. If all of the O2 is consumed, we'd get 2*1.153 or 2.31 moles of H2O. Convert to grams (2.31 moles*18 grams/mole) = 41.5 grams of water.
Bob
[Note that we can determine the amount of oxgen remaining. If all 1.153 moles of the oxygen were consumed, it would require twice that many moles of H2. That is 2.306 moles of hydrogen consumed, leaving 2.94 moles unreacted. That's 5.89 grams of the precious gas. A Waste!
Out of curiosity, the Law of Conservation of Mass tells us we should have the same total mass after the reaction. Let's see if that is true:
Initial (g): H2 O2 Final: H2O Leftover H2
10.5 36.9 41.5 5.89
Totals: Initial: 47.4 grams Final: 47.4 grams
Nice.
Bob
