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Can you help me balance this chemical equation? N3+H2 --> NH3thanks!
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High School Teacher
To balance chemical equations, you need to see what combination of numbers will allow all of the atoms' numbers to be the same across the reaction arrow.
You can start by adding coefficients to each term:
a*N3 + b*H2 --> c*NH3
You can actually now set up an equation to relate the coefficients to each other:
In terms of nitrogen:
3a = c (no b term because there is no nitrigen in hydrogen)
In terms of hydrogen:
2b = 3c (no a term because azide has no hydrogen!)
Well, these two equations give us a pretty nice relation in that 3a = c. We can substitute for c in the second equation using (3a) as a result.
2b = 3(3a)
2b = 9a
Now, remembering that a = 3c, we get a sort of symmetric equation:
2b = 9a = 3c = k
So, all we'd need to do is select a number that gives us the least-common-multiple for each number. This is the k that we're setting all three expresions to. In other words, for this equation, we'll need to find the LCM for 2, 3, and 9. This number ends up being 18, so we can solve for a, b, and c by setting up the three equations that result from our setting the symmetric equation to 18.
2b = 18
3c = 18
9a = 18
We can find fairly quickly that a = 2, b = 9, and c = 6
Now, we can just plug in those coefficients to find our balanced chemical equation:
2(N3) + 9(H2) --> 6NH3
I hope that helps!
Posted by txmedteach on February 1, 2012 at 2:07 AM (Answer #1)
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