Noob question- How to balance this equation? (1 Viewer)

[girlworld_club]

Butane + oxygen --> carbon dioxide + water

Please show full working, and how it eventually balances. Thanks!

btw. Butane is C4H10 (both sub scripts)

 

[ademb13]

first balance the carbon on both sides. Since there is 4 carbon on LHS, you put a 4 in front of the C02 on RHS.
Next balance the Hydrogen on both sides. Since there is 10 Hydrogen on LHS put a 5 in-front of the H2O on RHS.
Finally the oxygen, count the amount of oxygen on the RHS then put half of the number you count in-front of the oxygen on the LHS (as oxygen is 02).

you will get the hang of it

 

[ademb13]

so it becomes:
c4h10 + 13/2 o2 -----> 4Co2 + 5h2o

 

[girlworld_club]

so it becomes:
c4h10 + 13/2 o2 -----> 4Co2 + 5h2o

the answer at the back is 2c4H10 + 1302 ---> 8Co2 + 10 H20

 

[teeah]

I think that's pretty much the same thing, you're just mulitiplying it by 2 throughout so you don't have a fraction infront of the oxygen?

 

[skillstriker]

the answer at the back is 2c4H10 + 1302 ---> 8Co2 + 10 H20

just double what ademb13 said

 

[girlworld_club]

why do i end up getting

C4H10 + 13o2 ---> 4Co2 + 5H20

what am i doing wrong?

 

[skillstriker]

why do i end up getting

C4H10 + 13o2 ---> 4Co2 + 5H20

what am i doing wrong?

you didn't double the C4H10 and 5H20

 

[Shadowdude]

why do i end up getting

C4H10 + 13o2 ---> 4Co2 + 5H20

what am i doing wrong?

Show working.

 

[deswa1]

why do i end up getting

C4H10 + 13o2 ---> 4Co2 + 5H20

what am i doing wrong?

You have two times as many oxygens as you need on the LHS. Type up your working so we can see exactly where you made the mistake- remember to balance the oxygen last because its by itself.

 

[deswa1]

<a href="http://www.codecogs.com/eqnedit.php?latex=C_{4}H_{10}@plus;O_{2}->CO_{2}@plus;H_{2}O\\ C_{4}H_{10}@plus;O_{2}->4CO_{2}@plus;H_{2}O\\ C_{4}H_{10}@plus;O_{2}->4CO_{2}@plus;5H_{2}O\\ C_{4}H_{10}@plus;\frac{13}{2}O_{2}->4CO_{2}@plus;5H_{2}O\\ 2C_{4}H_{10}(g)@plus;13O_{2}_{(g)}->8CO_{2}_{(g)}@plus;10H_{2}O_{(l)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?C_{4}H_{10}+O_{2}->CO_{2}+H_{2}O\\ C_{4}H_{10}+O_{2}->4CO_{2}+H_{2}O\\ C_{4}H_{10}+O_{2}->4CO_{2}+5H_{2}O\\ C_{4}H_{10}+\frac{13}{2}O_{2}->4CO_{2}+5H_{2}O\\ 2C_{4}H_{10}(g)+13O_{2}_{(g)}->8CO_{2}_{(g)}+10H_{2}O_{(l)}" title="C_{4}H_{10}+O_{2}->CO_{2}+H_{2}O\\ C_{4}H_{10}+O_{2}->4CO_{2}+H_{2}O\\ C_{4}H_{10}+O_{2}->4CO_{2}+5H_{2}O\\ C_{4}H_{10}+\frac{13}{2}O_{2}->4CO_{2}+5H_{2}O\\ 2C_{4}H_{10}(g)+13O_{2}_{(g)}->8CO_{2}_{(g)}+10H_{2}O_{(l)}" /></a>

 

[ademb13]

all you had to do was x by 2 to get rid of the fraction