Posted by lyne on Friday, January 23, 2009 at 6:35pm.
#1. Write the equation and balance it.
N2 + 3H2 ==> 2NH3
initial concn.
N2 = 6 N2 molecules
H2 = 6 H2 molecules
NH3 = 0 molecules.
If it goes to completion, one of these must be the limiting reagent. How much NH3 would 6 molecules N2 produce? That will be
6 molecules N2 x (2 moles NH3/1 mole N2) = 12 molecules NH3.
How much NH3 would 6 molecules H2 produce?
6 molecules H2 x (2 moles NH3/3 moles H2) = 4 molecules NH3.
The smaller number is 4; therefore, H2 is the limiting reagent and 4 molecules NH3 will be produced. How much N2 will that use.
6 molecules H2 x (1 mole N2/3 moles H2) = 2 molecules N2.
Final mixture:
NH3 = 4 molecules
H2 = 6 molecules to start; we used 6 so we end up with 0 molecules H2.
N2 = We start with 6 molecules, we use 2 molecules N2 so we are left with 6-2=4.
Check my work.
If you wrote the balanced chemical equation you know the answer.
okay that makes perfect sense thanks :)
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