Asked by Chopsticks
We are on combustion reaction at the moment in chemistry, and I have to say, these reactions are like extremely hard to balance(well, just alot of erasing). It takes me a while to balance combustion reactions and I want to know if theres any tips in balancing them? Should I balance a certain element first to make it easier?
Answers
Answered by
DrBob222
Look at the products first to determine how much of the reactants you need.
I balance oxygen last because you can add (usually) exactly what you need as a single element on the left. Then if it comes out something and a half, multiply by 2.
I balance oxygen last because you can add (usually) exactly what you need as a single element on the left. Then if it comes out something and a half, multiply by 2.
Answered by
Chopsticks
Today, we balanced combustion reactions and I got numbers up to 62! Big numbers are intimidating to me :P. is it always going to be like that for combustion reactions?
Answered by
DrBob222
Most combustion reactions are small numbers.
C4H9 + O2 --> CO2 + H2O
I see the H9 on the left and H2 on the right and I know I have a problem right off the bat because there is no whole number I can multiply by to get from 2 to 9. So right off the bat I multiply C4H9 by 2
2C4H9 + O2 ==> CO2 + H2O
Now with 18 H on the left I can balance H on the right.
2C4H9 + O2 ==> CO2 + 9H2O
Now I can balance C.
2C4H9 + O2 ==>8CO2 + 9H2O
Now I count the O I have on the right. That's 16 from CO2 and 9 from H2O which makes 25, another odd number. Usually, what I do is say, ok, I need 25/2 O2 like this,
2C4H9 + (25/2)O2 ==> 8CO2 + 9H2O
Then I multiply everything through by 2 to get rid of the half so that 25/2 becomes 25.
4C4H9 + 25 O2 ==>16CO2 + 18H2O
Then I check it. ALWAYS check it.
16 C left and right.
36 H left and right.
50 O left and 50(32 from CO2 + 18 from H2O = 32+18 = 50).
One thing about balancing equations is that you can ALWAYS know if you are right by checking the equation when you are finished. I always check mine to make sure everything balances.
C4H9 + O2 --> CO2 + H2O
I see the H9 on the left and H2 on the right and I know I have a problem right off the bat because there is no whole number I can multiply by to get from 2 to 9. So right off the bat I multiply C4H9 by 2
2C4H9 + O2 ==> CO2 + H2O
Now with 18 H on the left I can balance H on the right.
2C4H9 + O2 ==> CO2 + 9H2O
Now I can balance C.
2C4H9 + O2 ==>8CO2 + 9H2O
Now I count the O I have on the right. That's 16 from CO2 and 9 from H2O which makes 25, another odd number. Usually, what I do is say, ok, I need 25/2 O2 like this,
2C4H9 + (25/2)O2 ==> 8CO2 + 9H2O
Then I multiply everything through by 2 to get rid of the half so that 25/2 becomes 25.
4C4H9 + 25 O2 ==>16CO2 + 18H2O
Then I check it. ALWAYS check it.
16 C left and right.
36 H left and right.
50 O left and 50(32 from CO2 + 18 from H2O = 32+18 = 50).
One thing about balancing equations is that you can ALWAYS know if you are right by checking the equation when you are finished. I always check mine to make sure everything balances.
Answered by
Chopsticks
Ah, another problem I find that I have is that sometimes, I use double the number im suppose to use, and even those they balance, its 2x what I need. Is there a way to check that? it doesnt happen often, but it does.
Answered by
DrBob222
Sometimes that happens to me, too. I just end up dividing everything by two and don't worry about it.
Answered by
H + O2
H20
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