8.40 grams of Hydrogen gas reacts with 48.2 grams of Oxygen gas.

Give the mass of the water vapor produced and the mass of the excess reactant left over.

Can somebody please explain stoichiometry to me? I've failed the test and the retest. I'm usually able to balance the equation, but after that I get stuck. I just don't get it. :(

What you have posted is a limiting reagent problem. Basically, you solve those by solving TWO simple stoichiometry problems, identifying the limiting reagent (it produces the smaller amount of product), and go from there.

Here is a stoichiometry problem with all of the steps. Just follow the steps. Post back if there is something you don't understand.
http://www.jiskha.com/science/chemistry/stoichiometry.html

Umm, I don't understand, how do I found out what the limiting reagent is?

Stoichiometry is a concept in chemistry that deals with the quantitative relationship between reactants and products in a chemical reaction. It allows you to calculate the amount of reactants needed and the amount of products produced in a given reaction.

To understand stoichiometry, it is important to start with a balanced chemical equation. The balanced equation represents the proportions in which reactants combine and products are formed. For example, the balanced equation for the reaction between hydrogen gas (H2) and oxygen gas (O2) to form water (H2O) is:

2H2 + O2 -> 2H2O

The coefficients in the balanced equation (the numbers in front of the reactants and products) indicate the stoichiometric ratio between them. In this case, it means that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.

Now, let's move on to solve the problem you have mentioned.

1. Calculate the number of moles of hydrogen and oxygen:
To find the number of moles, you use the molar mass of each substance. The molar mass of hydrogen is 2 g/mol, and the molar mass of oxygen is 32 g/mol. Therefore,

Number of moles of hydrogen = mass of hydrogen / molar mass of hydrogen
= 8.40 g / 2 g/mol
= 4.20 moles

Number of moles of oxygen = mass of oxygen / molar mass of oxygen
= 48.2 g / 32 g/mol
= 1.51 moles

2. Determine the limiting and excess reactants:
The limiting reactant is the one that is completely consumed in the reaction and limits the amount of product that can be formed. To find the limiting reactant, you compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.

In this case, the stoichiometric ratio tells us that 2 moles of hydrogen react with 1 mole of oxygen. Since we have 4.20 moles of hydrogen and 1.51 moles of oxygen, the oxygen is the limiting reactant.

3. Calculate the amount of water vapor produced:
From the stoichiometric ratio, we know that 1 mole of oxygen reacts to produce 2 moles of water. Therefore, the number of moles of water produced is:

Number of moles of water = (number of moles of oxygen) x (moles of water per mole of oxygen)
= 1.51 moles x 2 moles of water / 1 mole of oxygen
= 3.02 moles

To find the mass of the water vapor produced, you multiply the number of moles by the molar mass of water (18 g/mol):

Mass of water vapor produced = (number of moles of water) x (molar mass of water)
= 3.02 moles x 18 g/mol
= 54.36 grams

Therefore, the mass of water vapor produced is 54.36 grams.

4. Determine the mass of the excess reactant left over:
Since the oxygen is the limiting reactant, all of the hydrogen will react completely. Therefore, there will be no excess hydrogen left over. However, if there was an excess reactant, you could calculate the mass of the excess reactant left over by subtracting the mass of the limiting reactant consumed from the total mass of the excess reactant.

In this case, the mass of the excess reactant left over (oxygen) would be 48.2 grams - (mass of oxygen used = (1.51 moles x 32 g/mol) = 48.32 grams = 0.12 grams *Notice the mistake when they subtracted

I hope this explanation helps you understand stoichiometry better. If you have any further questions or need clarification, feel free to ask!