Propane, c3hs. is flammable and used in rural areas where natural gas is not available. what is the coefficient of oxygen in the balanced equation for the combustion of propane? _c3 hp)* 026) > _co24) _h206) a) 1 b) 5 c) 10 d) 7 e) none of the above

The coefficient of oxygen in this balanced reaction is 5 ( option B)

Explanation:

Propane = C3H8

The process of combustion = adding oxygen (O2)

In this process propane will be consumed and CO2 and H2O will be produced.

C3H8 + O2 → CO2 + H2O

First we'll balance on both sides all H-atoms: left we have 8x H and on the right side we have 2x H so we have to multiply the right side ( H2O) by 4.

This gives us:

C3H8 + O2 → CO2 + 4H2O

Now we have on both side 8x H

On the left side we have 3x C, and 1x C on the right side. So we have to multiply CO2 on the right side by 3

This gives us: C3H8 + O2 → 3 CO2 + 4 H2O

On the left side we have 2x O but on the right side we have 10x O

We have to multiply O2 on the left side by 5, to equal this.

So finally, we'll have: C3H8 + 5 O2  → 3 CO2 + 4 H2O

We can see that the coefficient of oxygen in this balanced reaction is 5 ( option B)

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).