# Chemistry

posted by
**chrissy** on
.

propane(C^3H^8)is used as a fuel. Propane is a type of hydrocarbon. Write a balanced equation for the combustion of propane in oxygen. Assume that the combustion is complete.

Hydrocarbons (C and H compounds of which propane is one) burn in oxygen to produce CO2 and H2O. Your formula is incorrect. It should be written as

C_{3}H_{8} but since it is so difficult to write subscripts on these boards we usually write it as C3H8 with the understanding that 3 and 8 are subscripts. If we wish to show superscripts, THEN we use the caret as you did. A number in front of a compound, as in 3CO2, is a coefficient and not a subscript.

C3H8 + O2 ==> CO2 + H2O.

I will leave it for you to balance because you need the practice. BUT, let me give you some hints.

See on the left we have 3 C in C3H8. So make the C atoms on the right 3 (Remember not to change any of the subscripts). Then look at the H on the left and see 8 H atoms. Make 8 on the right from H2O. Then count up oxygen atoms needed on the right and make O atoms on the left that number.

Can you please check to see if I have balaced this right then:

C3H8+O2--->3CO2+8H2O

I think it is all right but maybe when I did the oxygen atoms.

No. You should check it to see. Here is what I see.

3c on the left and 3C on the right. ok.

8H on the left and 16 on the right(from 8H2O--8x2=16). not ok.

2 O on the left and 14 O on the right(6 from 3CO2 and 8 from 8H2O.)not ok.

Here is how I do it.

C3H8+O2--->CO2+H2O

I look at C on the left and I have 3. Therefore, I place a 3 in front of the CO2 which you have done. good.

C3H8+O2--->3CO2+H2O

Then look at H on the left and there are 8 so I place a 4 in front of H2O to make 8 on the right.

C3H8+O2--->3CO2+4H2O

Those two are fairly easy because all of the H is in one place on both sides and all of the C is in one place on both sides. Not so with oxygen. It is in one place on the left and two places on the right. The easy way to do this is to count how much we have on the right, then make the left side match. I see 10 O atoms on the right (6 from 3CO2 and 4 from 4H2O to make 6 + 4 = 10). Therefore, I must have 10 on the left. Thus, I place a 5 in front of O2 which will give me 10.

C3H8 + 5O2 --> 3CO2 + 4H2O

I ALWAYS check them when I finish to guard against making an error. I see 3C on the left (C3H8) and 3C on the right (3CO2). ok.

I see 8 H on the left (C3H8) and 8 on the right(4H2O). ok.

I see 10 O on the left (from 5O2) and 10 on the right (6 from 3CO2 and 4 from 4H2O). ok.

For whatever it is worth I think you are making this far too difficult. I know I have had much more practice than you BUT these are fairly simple to do. Just keep practicing. You're getting better.