How do you solve stoichiometry problems?

Answer 1

To solve stoichiometry problems, you must first do two very important things.

1) Write a balanced equation for the reaction.
2) Convert all amounts of products and/or reactants in the question into moles.
To find out how to do both of these see the Related Questions links to the left of this answer. (Note that if the question involves gases, and the amount of gas is given as a volume, you need to use the Ideal Gas Law. How to do that is also listed under the Related Questions).
Once all quantities have been converted to moles and you have a balanced reaction, you are ready to actually use stoichiometry. The idea of stoichiometry is really quite simple. The coefficients (or numbers) in front of each reactant and product in the balanced chemical reaction tells you the ratio of how much of each you will react/produce. Let's take a simple example: the reaction of hydrogen gas (H2) with oxygen gas (O2) to form water (H2O). 2H2 + O2 ---> 2H2O In the balanced reaction, there is a 2 in front of the H2, and although nothing is written, that means that there is really 1 in front of O2, and a 2 in front of H2O. Stoichiometry tells us that because of the way the numbers in balanced reaction came out, we need two molecules of H2 to react with each one molecule of O2, and also, that this will form 2 molecules of water. I can say the same thing using moles: For each mole of O2 I react, I need 2 moles of H2, and I will produce 2 moles of O2. All I'm using is the ratio of the coefficients from the balanced reaction. That is stoichiometry! So, using this I can say things like: -- If I reaction 0.5 moles of O2 completely, I will make 1 mole of water
-- If I made 4 moles of water, then I consumed 4 moles of H2 and 2 moles of O2.
-- If I want to completely react 5 moles of H2, then I need 2.5 moles of O2 (and I will get 5 moles of H2O). Makes sense? The ratio of H2 to O2 to H2O is 2:1:2, and that always holds for this reaction.
Now let's make it a bit more complicated.
--Problem: If I burn 10 grams of methane, how many grams of CO2 will be produced?
-- Answer: As I said, to solve this problem we need two things: a balanced reaction, and to convert all quantities to moles. First, let's write a balanced reaction. In this reaction, methane gas (CH4) gets burned in oxygen (O2) to form carbon dioxide (CO2) and water vapor (H2O). The balanced reaction is:
CH4 + 2O2 --> CO2 + 2H2O
Notice the ratio is now 1:2:1:2 (reading the reaction from left to right). So that tells me for every mole of CH4 I react, I need 2 moles of O2, and I will get out 1 mole of CO2 and 2 moles of H2O.
Now, I need to convert the 10 grams of methane into moles, because stoichiometry only works for moles and NOT grams! So I use the molar mass of CH4, which is 12.011+4*1.0079 = 16.0426 grams per mole. So to convert to moles, I just divide: 10 grams ÷ 16.0426 gram/mole = 0.6233 moles From the stoichiometry, I now know that if I react 0.6233 moles of methane, I will need twice that many moles of oxygen, or 1.2467 moles O2, and I will get 0.6233 moles of CO2 and 1.2467 moles of H2O as products. But the question asked for a number of grams of CO2, not moles. So once I'm done using stoichiometry, I convert back to grams, now using the molar mass of CO2 (which is 12.011 + 2*15.999 = 44.009 g/mol). So 0.6233 moles of O2 is: 0.6233 moles * 44.009 grams/mole = 27.4308 grams of CO2 Notice that the ratio of moles of CH4 to CO2 is 1 to 1, but the ratio of the weights is totally different. Remember, STOICHIOMETRY ONLY WORKS ON MOLES! I can also find how many grams of water I'll produce, just for fun! The molar mass of water is 15.999 + 2*1.0079 = 18.0148 g/mol. So to convert 1.2467 moles H2O to grams: 1.2467 moles * 18.0148 gram/mole = 22.4591 grams of H2O
to answer stoichiometry problems there are several ways.

one way is the mole concept(mol)

-always remember Avogadro's number which is 6.02x10^23.

eg. how many atoms are there in 1 mol of glucose(C6H12O6)?

1 mol C6H13O6 x 6.02x10^23= 6.02x19^23

1 mol C6H12O6

cancel 1 mol C6H12O6 leaving the answer.