Question

Write the complete balanced equation for the following reaction:

C5H9O + O2 yields CO2 + H2O

please help, i'm lost :/

Answer 1

http://answers.yahoo.com/question/index?qid=20070821110959AAEaSXf

Answer 2

well i need someone to explain it for me

Answer 3

Okay.
Thomaster is great at that stuff :D

Answer 4

\(\large\sf C_5H_9O+O_2 \to CO_2+H_2O\)

First write down the number of atoms you have on teach side of the equation
For this one, that would be:
left side: 5x C, 9x H and 3x O
right side: 1x C, 2x H and 3x O

You start with the carbon atoms, add coefficients to a place where carbon is with the least other atoms. That would be CO2

To equal the C's you'd need to add coefficient 5 to the CO2 on the right side, so:
\(\large\sf C_5H_9O+O_2 \to 5CO_2+H_2O\)

Now the C's are equal but the rest isn't
Left side: 5x C, 9x H and 3x O
Right side: 5x C, 2x H and 11x O

Follwing it so far? :P

Answer 5

@dumbsearch2 please don't answer a question with another answer if you don't know the material yourself! You could be giving them the wrong answer. So STOP!

Answer 6

ok so do i put 4.5 in front of H2O??

Answer 7

\[\huge STOP\]

Answer 8

._.

Answer 9

nonono no decimals in the coefficients :P

next we're going to balance hydrogen.
\(\large\sf C_5H_9O+O_2 \to 5CO_2+H_2O\)
Left side: 5x C, 9x H and 3x O
Right side: 5x C, 2x H and 11x O

We cannot add 4.5 to H2O so we add a 2 before the C5H9O
Since we double the C4H9O, we also need to double the CO2 to keep it balanced.
\(\large\sf 2C_5H_9O+O_2 \to 10CO_2+H_2O\)
Left side: 10x C, 18x H and 4x O
Right side: 10x C, 2x H and 21x O

To balance the hydrogen, add coefficient 9 before H2O (because we have 18 H's on the right and H2O contains 2 H's)

\(\large\sf 2C_5H_9O+O_2 \to 10CO_2+9H_2O\)
Left side: 10x C, 18x H and 4x O
Right side: 10x C, 18x H and 29x O

Last step is balancing oxygen

Answer 10

this is really confusing :c

Answer 11

\(\large\sf 2C_5H_9O+O_2 \to 10CO_2+9H_2O\)

we have 4 O's on the left and 29 O's on the right
29 is an odd number, needs to be an even number so we multiply everything by 2

\(\large\sf 4C_5H_9O+2O_2 \to 20CO_2+18H_2O\)
Left side: 20x C, 36x H and 8x O
Right side: 20x C, 36x H and 58x O

Now we can balance the O2
To get the O's on the left side to 58, we need to change the coefficient before O2 on the left side to 27 (27*2=54 and the 4xO from C4H9O, that makes 58 total)

So the final balanced equation is:
\(\large\sf 4C_5H_9O+27O_2 \to 20CO_2+18H_2O\)

Answer 12

Still confused?

Answer 13

how did you know to double C4H9O and CO2

Answer 14

Because there is a 9 in C4H9O, on the left side is H2O, to balance that we need to add 4,5 as coefficient before the H2O. we can only add whole numbers.

Answer 15

So by multiplying it by 2 you'd get 18 and 2, to balance that you change the 2 in 9

Answer 16

hmm..ok..thanks

Answer 17

Oke, tell me if you don't understand something in the explanation :)

Answer 18

can you tell me if this is right?

Answer 19

nvm, i got it :]