How would you balance this equation: C_2H_6+O_2 -> CO_2+H_2O?

Nov 21, 2015

${\text{2C"_2"H"_6" + 7O}}_{2}$$\rightarrow$$\text{4CO"_2" + 6H"_2"O}$

Explanation:

${\text{C"_2"H"_6" + O}}_{2}$$\rightarrow$$\text{CO"_2" + H"_2"O}$

Balance H first.

There are 6 H atoms on the left and 2 on the right. Place a coefficient of 3 in front of the $\text{H"_2"O}$.

${\text{C"_2"H"_6" + O}}_{2}$$\rightarrow$$\text{CO"_2" + 3H"_2"O}$

Balance C next.

There are 2 C atoms on the left and 1 C atom on the right. Add a coefficient of 2 in front of the $\text{CO"_2}$.

${\text{C"_2"H"_6" + O}}_{2}$$\rightarrow$$\text{2CO"_2" + 3H"_2"O}$

Balance O last.

There are 2 O atoms on the left and 7 O atoms on the right.

Place a coefficient of 7/2 in front of $\text{O"_2}$ on the left.

${\text{C"_2"H"_6" + "7/2"O}}_{2}$$\rightarrow$$\text{2CO"_2" + 3H"_2"O}$

All of the coefficients in a chemical equation must be whole numbers. To change $\frac{7}{2}$ to a whole number, multiply all coefficients times 2.

${\text{2C"_2"H"_6" + 7O}}_{2}$$\rightarrow$$\text{4CO"_2" + 6H"_2"O}$

There are now 4 C atoms on both sides, 12 H atoms on both sides, and 14 O atoms on both sides. The equation is balanced.