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

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Answer 1 · 2015-11-21T17:40:48

#"2C"_2"H"_6" + 7O"_2##rarr##"4CO"_2" + 6H"_2"O"#

#"C"_2"H"_6" + O"_2##rarr##"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 #"H"_2"O"#.

#"C"_2"H"_6" + O"_2##rarr##"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 #"CO"_2"#.

#"C"_2"H"_6" + O"_2##rarr##"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 #"O"_2"# on the left.

#"C"_2"H"_6" + "7/2"O"_2##rarr##"2CO"_2" + 3H"_2"O"#

All of the coefficients in a chemical equation must be whole numbers. To change #7/2# to a whole number, multiply all coefficients times 2.

#"2C"_2"H"_6" + 7O"_2##rarr##"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.