| If you can do these problems, stoichiometery should not be much of a problem as you are essentially performing the same math.
In this stage of the course you are most likely dealing with the 4 basic types of stoichiometry problems
1. Mole-Mole (given moles and converting to moles) 1 conversion factor
2. Mole-Mass (given moles and converting to mass) 2 conversion factors
3. Mass-Mole (given mass and converting to moles) 2 conversion factors
4. Mass-Mass (given mass and converting to mass) 3 conversion factors
In each of these cases you MUST have a correctly written, balanced equation in order to solve correctly.
Each type of conversion involves the use of one or more conversion factors. The number of conversion factors needed for each type is indicated above. (This is the way I teach it.)
In each case you must begin with the given value...
For number 1: Mole-Mole
(moles of given) x ((moles needed / moles given))
The information in the red parentheses comes from the balanced equation. You will use the coefficients in front of the needed (what the problem is asking for) and the given to fill in here. This is called the molar ratio. It will always be a whole number.
Example:
2H2 + O2 --> 2H2O
Given 8.3 moles hydrogen, how many moles of water can be produced?
8.3 mol H2 x (2 mol H2O / 2 mol H2)
Notice the H2 will cancel leaving the H2O. Moles needed goes on top. What the question is asking for goes on top. The question asks how much water so water is needed and it goes on top. We are given a value for H2 so that goes on the bottom. This information comes from the balanced equation only. In this reaction, there is always a ratio of 2 mol H2 to 2 moles of water.
Answer: 8.3 mol H2O
2. Mole-Mass
Use the same setup as above but add one more conversion factor. Convert to mass just like we did in your other question.
Imagine the above example altered slightly...
Given 8.3 moles hydrogen, what mass of water can be produced?
(moles of given) x (moles needed / moles given) x ((molar mass of needed / 1 mol needed))
8.3 mol H2 x (2 mol H2O / 2 mol H2) x (18.02g H2O / 1 mol H2O)
Answer: 149.57g H2O
3. Mass-Mole
Go in the reverse order...
How many moles of H2 are needed to produce 149.57g H2O?
mass of given x (1 mol given / molar mass of given) x (moles needed / moles given)
149.57g H2O x (1 mol H2O / 18.02g H2O) x (2 mol H2 / 2 mol H2O)
Notice the conversion factors are flipped from the other example. That is because the needed and given are different now but the rules stay essentially the same.
Answer: 8.3 mol H2
4. Mass-Mass
Take the last example and add one more conversion factor...
What mass of H2 is needed to produce 149.57g H2O?
mass of given x (1 mol given / molar mass of given) x (moles needed / moles given) x ((molar mass of needed / 1 mol needed))
149.57g H2O x (1 mol H2O / 18.02g H2O) x (2 mol H2 / 2 mol H2O) x ((2.02g H2 / 1 mol H2))
Answer: 16.77g H2
The fence is just a fancy way of setting up the conversion factors since we are just multiplying a bunch of fractions...
|-|x|-|x|-|x|-|
Stick with the layout rules I gave you and you should be OK.
Let me know if you need anything else.
Good Luck! | 
Linear Mode
