"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?

2 Answers
Mar 28, 2018

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.

Mar 28, 2018

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)