H. 21

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Maya Khoury
Posts: 29
Joined: Fri Apr 06, 2018 11:03 am

H. 21

Postby Maya Khoury » Sun Apr 08, 2018 3:25 pm

Problem H.21 took me a while to figure out; It asks you to balance the equation of C₁₀H₁₅N + O₂ --> CO₂ + H₂O + CH₄N₂O

How would you suggest we start the problem? Is it that we should begin with the element that appears the least frequent in the equation, for example with Nitrogen, and then move up from there?

I got the correct answer it just took me a while to know how to start.

thank you!

Chem_Mod
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Re: H. 21

Postby Chem_Mod » Sun Apr 08, 2018 3:47 pm

When you encounter a problem like this, a good method to do is to solve the algebraic way where you set up a system of equations where you balance all the elements and then try to solve for the coefficients.

AshleyLamba1H
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Joined: Fri Apr 06, 2018 11:03 am

Re: H. 21

Postby AshleyLamba1H » Sun Apr 08, 2018 6:23 pm

You should set the problem up so you are solving for a system of equations:
C: 10a+ 0b = 1c + 0d + 1e
H: 15h + 0b = 0c + 2d + 4e
N: 1a + 0b = 0c + 0d + 2e
O: 0a + 2b = 2c + 1d + 1e

Solve for each variable.

Julia Shapero 1E
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Joined: Fri Apr 06, 2018 11:05 am

Re: H. 21

Postby Julia Shapero 1E » Sun Apr 08, 2018 11:57 pm

So I'm still a bit confused on how this would work. Would someone be able to walk me through it? Also does this work for balancing any chemical equation?

Chem_Mod
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Re: H. 21

Postby Chem_Mod » Thu Jun 14, 2018 4:46 pm

C: 10a+ 0b = 1c + 0d + 1e
This corresponds to the number of carbons in each compound of the equation, with the five coefficients we're trying to solve for being a, b, c, d, and e.

This will work for balancing any equation, although sometimes the other way is faster


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