Given 6.0 mol of N2 are mixed with 12.0 mol of H2. Equation: N2 + 3H2-- 2NH3

Which chemical is in excess? What is the excess in moles?
Theoretically, how many moles of NH3 will be produced?
If the percentage yield of NH3 is 80%, how many moles of NH3 are actually produced?

Question

Given 6.0 mol of N2 are mixed with 12.0 mol of H2. Equation: N2 + 3H2--> 2NH3

Which chemical is in excess? What is the excess in moles?
Theoretically, how many moles of NH3 will be produced?
If the percentage yield of NH3 is 80%, how many moles of NH3 are actually produced?

anonymous · Answer

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1 mole of N2 reacts completely with 3 moles of H2. So, if you are given 6  moles of N2, they would react completely with 18 moles of H2, but you are given only 12 moles H2.
So H2 will get completely utilized but only some of N2 will get utilized in the reaction. Hence the chemical in excess(called excess reagent) is N2. 
Now,
3 moles of H2 react completely with--------- 1 mole N2
12 moles of H2 will react completely with-----1*12/3=4 mole N2
So only 4 moles of N2 react and the rest(6-4=2 moles) remains unreacted.