Question

`"C"_2"H"_(4("g")) + "O"_"2(g)" → "CO"_"2(g)" + "H"_2"O"_"(g)"` What is the coefficient of `"O"_2` in the balanced equation?

Answer 1

`3`

Explanation:

`"C"_2"H"_(4("g")) + "O"_"2(g)" → "CO"_"2(g)" + "H"_2"O"_"(g)"`

There are `2` carbons in reactant side and `1` carbon in products. So, add `2` before `"CO"_2`

`"C"_2"H"_(4("g")) + "O"_"2(g)" → 2"CO"_"2(g)" + "H"_2"O"_"(g)"`

There are `4` Hydrogens in reactant side and `2` Hydrogens in products. So, add `2` before `"H"_2"O"`

`"C"_2"H"_(4("g")) + "O"_"2(g)" → 2"CO"_"2(g)" + 2"H"_2"O"_"(g)"`

There are `2` oxygen in reactant side and `6` oxygens in products. So, add `3` before `"O"_2`

`"C"_2"H"_(4("g")) + 3"O"_"2(g)" → 2"CO"_"2(g)" + 2"H"_2"O"_"(g)"`

This is the balanced chemical equation. Coefficient of `"O"_2` as seen is `3`.

Answer 2

Oxygen coefficient = 3

Explanation:

STEPS:

Balance the number of CARBONS; 2.

Next, balance the number of HYDROGENS; 2 times 2.

Finally, balance the OXYGEN atoms created using the new coefficients; 6 total (3 times 2).

The balance coefficients equation is:

`C_2H_4 (g) + color(red)3 O_2 (g)` `"_----->` `color(blue)2CO_2 (g) + color(blue)2H_2 O (g)`