# Calculating moles from a balanced equation, together with enthalpy changes

1.Given the following reaction:
4 KO2 + 2 H2O + 4 CO2 ==> 4 KHCO3 + 3 O2

How many moles of O2 will be produced if 2.18 moles of CO2 react completely?

2.4NH3(g) + 3O2(g)  2N2(g) + 6H2O(g)

The enthalpy of the reaction DH = -1267 kJ. What is the enthalpy change (in kJ) when 3 grams of ammonia are burned?

3. 4 P + 5 O2 ==> P4O10

If you have 28.20 grams of P and 75.05 grams of O2, how many moles of each compound do you have present?

Okay, in all of these problems it is important to realize what information is being provided in the balanced equations for each of these reactions. Let’s review by looking at an example of another reaction first.

3H2SO4 + 2Al(OH)3  ===>  Al2(SO4)3 + 6H2O

Here we have sulfuric acid reacting with aluminum hydroxide forming the salt aluminum sulfate and water as products. Notice that we have to types of numbers here. We have the leading coefficients 3, 2, 1 (it is implied that if the coefficient is not written it is 1) and 6. These represent how many “parts” of each product or reactant are involved in the reaction. These parts can be individual molecules of the chemical or we can think of them in terms of number of moles of molecules reacting and being produced. With these we can establish any ratio between any two chemicals, whether they are reactants or products.

In this example we can say that for every 3 parts of sulfuric acid reacted, 1 part aluminum hydroxide is produced or we could also say that for every 3 parts of sulfuric acid there are 6 parts of water produced. Thinking of these “parts” as moles now we can say for every 3 moles of sulfuric acid we get 1 mole of aluminum sulfate and 6 moles of water. We might want to look at it from the aluminum standpoint. We could say that for every 2 moles of aluminum reacted we get 6 moles of water. Again, we can set these types of ratios up for any two compounds in a reaction but I think you get the idea now.

So now, suppose I was to ask you how many moles of water will be produced if 4.6 moles of aluminum hydroxide were completely reacted. We can now use these ratios to solve these types of problems using a method called dimensional analysis.

Now we want to establish the ratio of moles of aluminum hydroxide to moles of water in the reaction. We see from our coefficients that for every 2 moles of aluminum hydroxide 6 moles of water are produced. Now we can set up our dimensional analysis:

(4.6 mol Al(OH)3 / 1) * (6 mol H2O / 2 mol Al(OH)3) = 13.8 mol H2O produced

These are the same methods you want to use to solve your questions here. In your first problem, do what we did in this example but this time your ratios are going to involve O2 and CO2 but the method you will use is exactly the same.

Same method is going to be used for your enthalpy change. The ?H value should be expressed as KJ / mol rxn. Again look at the leading coefficients and determine how many moles are involved. Next, convert grams of ammonia to moles and then multiply by your ratio of KJ / mol of NH3 in the rxn.

For your third question, you are told the amounts (in grams) of each reactant but you want them to be in moles so convert grams to moles. This problem is what is referred to as a limiting reactant problem because one of the reactants will limit how much of the product can actually be produced and the other reactant that is not the limiting reactant will have excess amount left over and unreacted. Do you understand how to solve limiting reactant problems? They’re really just another type of problem that uses stoichiometry.

If you need more help please come back and we can continue this discussion but I think what we have done here should get you on the right track. Please feel free to ask more questions if anything here was not clear.