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

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

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Answer 1 · 2014-05-28T23:18:11

The balanced equation is $C_{7} H_{16} + {11O}_{2} \to {7CO}_{2} + {8H}_{2} O$ $"C"_7"H"_16 + "11O"_2 → "7CO"_2 + "8H"_2"O$ .

Explanation:

You follow a systematic procedure to balance the equation.

Start with the unbalanced equation:

$C_{7} H_{16} + O_{2} \to {CO}_{2} + H_{2} O$ $"C"_7"H"_16 + "O"_2 → "CO"_2 + "H"_2"O"$

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

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

The most complicated formula looks like $C_{7} H_{16}$ $"C"_7"H"_16$ . We put a $1$ $1$ in front of it to remind ourselves that the coefficient is now fixed.

We start with

$1 C_{7} H_{16} + O_{2} \to {CO}_{2} + H_{2} O$ $color(red)(1)"C"_7"H"_16 + "O"_2 → "CO"_2 + "H"_2"O"$

Balance $C$ $"C"$ :

We have fixed 7 $C$ $"C"$ atoms on the left-hand side, so we need 7 $C$ $"C"$ atoms on the right-hand side. We put a $7$ $7$ in front of the ${CO}_{2}$ $"CO"_2$ .

$1 C_{7} H_{16} + O_{2} \to 7 {CO}_{2} + H_{2} O$ $color(red)(1)"C"_7"H"_16 + "O"_2 → color(blue)(7)"CO"_2 + "H"_2"O"$

Balance $O$ $"O"$ :

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

Balance $H$ $"H"$ :

We have fixed 16 $H$ $"H"$ atoms on the left-hand side, so we need 16 $H$ $"H"$ atoms on the right-hand side. We put an $8$ $8$ in front of the $H_{2} O$ $"H"_2"O"$ .

$1 C_{7} H_{16} + O_{2} \to 7 {CO}_{2} + 8 H_{2} O$ $color(red)(1)"C"_7"H"_16 + "O"_2 → color(blue)(7)"CO"_2 + color(green)(8)"H"_2"O"$

Now we can balance $O$ $"O"$ :

We have fixed 22 $O$ $"O"$ atoms on the right-hand side: 14 from the ${CO}_{2}$ $"CO"_2$ and 8 from the $H_{2} O$ $"H"_2"O"$ . We put an $11$ $11$ in front of the $O_{2}$ $"O"_2$ .

$1 C_{7} H_{16} + 11 O_{2} \to 7 {CO}_{2} + 8 H_{2} O$ $color(red)(1)"C"_7"H"_16 + color(teal)(11)"O"_2 → color(blue)(7)"CO"_2 + color(green)(8)"H"_2"O"$

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

Let’s check:

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

All atoms balance. The balanced equation is

$C_{7} H_{16} + 11 O_{2} \to 7 {CO}_{2} + 8 H_{2} O$ $"C"_7"H"_16 + 11"O"_2 → 7"CO"_2 + 8"H"_2"O"$

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

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