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QUESTION

What is the balanced equation for heptane (C7H16) burning in oxygen to make carbon dioxide and water?

The balanced equation is "C"_7"H"_16 + "11O"_2 → "7CO"_2 + "8H"_2"O.

You follow a systematic procedure to balance the equation.

"C"_7"H"_16 + "O"_2 → "CO"_2 + "H"_2"O"

A method that often works is to balance everything other than "O" and "H" first, then balance "O", and finally balance "H".

Another useful procedure is to start with what looks like the most complicated formula.

The most complicated formula looks like "C"_7"H"_16. We put a 1 in front of it to remind ourselves that the coefficient is now fixed.

color(red)(1)"C"_7"H"_16 + "O"_2 → "CO"_2 + "H"_2"O"

Balance "C":

We have fixed 7 "C" atoms on the left-hand side, so we need 7 "C" atoms on the right-hand side. We put a 7 in front of the "CO"_2.

color(red)(1)"C"_7"H"_16 + "O"_2 → color(blue)(7)"CO"_2 + "H"_2"O"

Balance "O":

We can't balance "O" yet because we have two formulas that contain "O" and lack coefficients. So we balance "H" instead.

Balance "H":

We have fixed 16 "H" atoms on the left-hand side, so we need 16 "H" atoms on the right-hand side. We put an 8 in front of the "H"_2"O".

color(red)(1)"C"_7"H"_16 + "O"_2 → color(blue)(7)"CO"_2 + color(green)(8)"H"_2"O"

Now we can balance "O":

We have fixed 22 "O" atoms on the right-hand side: 14 from the "CO"_2 and 8 from the "H"_2"O". We put an 11 in front of the "O"_2.

color(red)(1)"C"_7"H"_16 + color(teal)(11)"O"_2 → color(blue)(7)"CO"_2 + color(green)(8)"H"_2"O"

Every formula now has a fixed coefficient. We should have a balanced equation.

Let’s check:

color(white)(m)"Element"color(white)(m)"Left-hand side"color(white)(m)"Right-hand side" color(white)(mmll)"C"color(white)(mmmmml)7color(white)(mmmmmmmmll)7 color(white)(mmll)"H"color(white)(mmmmll)16color(white)(mmmmmmmm)16 color(white)(mmll)"O"color(white)(mmmmll)22color(white)(mmmmmmmm)22

All atoms balance. The balanced equation is

"C"_7"H"_16 + 11"O"_2 → 7"CO"_2 + 8"H"_2"O"