Hello, Leslie,
Your sentence is missing something, so I changed it to what I think you may be asking.
How many moles of ammonia are produced when 13 moles of nitrogen reacts with an excess of hydrogen?
If this is the question, here is how you can solve for it:
We need a balanced equation for the reaction in question:
N2 + 3H2 = 2NH3
There are 2 N and 6 H on both sides, so it is balanced. (By convention, we must start with the diatomic forms of both hydrogen and nitrogen).
This equation tells us either 1) one molecule of nitrogen reacts with 3 molecules of hydrogen to form 2 molecules of ammonia, or 2) one mole of N2 reacts with 3 moles of H2 to form 2 moles of NH3.
It is convenient for us to use the second interpretation, since we want to find moles of ammonia produced. If one mole of N2 gives 2 moles of NH3, then 13 moles of N2 must give us (2*13) moles of NH3. The balanced equation gives us the molar ratios of the reactants and products. In this case, the ratio we are interested in is (2 moles NH3)/(1 mole N2) or 2 (moles NH3)/(mole N2).
So we would expect 26 moles of ammonia to be produced from 13 moles of nitrogen, given that we have an excess of H2. If we wanted to use only the exact amount of H2 required, we can see that the molar ratio of H2/N2 is 3(moles H2/moles N2). So a crafty engineer would add only (3moles H2/moles N2)*(13 moles N2 ) = 36 moles of H2 and save the excess for floating and igniting balloons for his wife's birthday.
Bob
