Question #7ba58

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Question #7ba58

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Answer 1 · 2017-04-30T21:56:22

Warning! Long Answer. You follow a systematic procedure to balance the equation.

Explanation:

Start with the unbalanced equation:

$C_{8} H_{18} + O_{2} \to {CO}_{2} + H_{2} O$ $"C"_8"H"_18 + "O"_2 → "CO"_2 + "H"_2"O"$

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 $C_{8} H_{18}$ $"C"_8"H"_18$ . We put a 1 in front of it to remind ourselves that the number is now fixed.

$1 C_{8} H_{18} + O_{2} \to {CO}_{2} + H_{2} O$ $color(red)(1)"C"_8"H"_18 + "O"_2 → "CO"_2 + "H"_2"O"$

Balance $C$ $"C"$ :

We have fixed $8C$ $"8C"$ on the left, so we need $8 C$ $"8 C"$ on the right. Put an 8 in front of the ${CO}_{2}$ $"CO"_2$ .

$1 C_{8} H_{18} + O_{2} \to 8 {CO}_{2} + H_{2} O$ $color(red)(1)"C"_8"H"_18 + "O"_2 → color(blue)(8)"CO"_2 + "H"_2"O"$

Balance $O$ $"O"$ :

We can't balance $O$ $"O"$ because we have two formulas containing $O$ $"O"$ that have no coefficients.

Balance $H$ $"H"$ :

We have fixed $18 H$ $"18 H"$ on the left, so we need $18 H$ $"18 H"$ on the right. Put a 9 in front of $H_{2} O$ $"H"_2"O"$ .

$1 C_{8} H_{18} + O_{2} \to 8 {CO}_{2} + 9 H_{2} O$ $color(red)(1)"C"_8"H"_18 + "O"_2 → color(blue)(8)"CO"_2 + color(orange)(9)"H"_2"O"$

Balance $O$ $"O"$ :

We have fixed $25 O$ $"25 O"$ on the right, so we need 25 $O$ $"O"$ on the left.

Oops! That means we would need ${12.5 O}_{2}$ $"12.5 O"_2$ on the left.

My instructor never let me use fractions when balancing equations.

The solution: Multiply every coefficient by 2.

$2 C_{8} H_{18} + O_{2} \to 16 {CO}_{2} + 18 H_{2} O$ $color(red)(2)"C"_8"H"_18 + "O"_2 → color(blue)(16)"CO"_2 + color(orange)(18)"H"_2"O"$

Now, we have $50 O$ $"50 O"$ on the right, so we can use ${25 O}_{2}$ $"25 O"_2$ on the left.

$2 C_{8} H_{18} + 25 O_{2} \to 16 {CO}_{2} + 18 H_{2} O$ $color(red)(2)"C"_8"H"_18 + color(purple)(25)"O"_2 → color(blue)(16)"CO"_2 + color(orange)(18)"H"_2"O"$

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

Let’s check:

$Atom m Left hand side m Right hand side$ $bb("Atom"color(white)(m)"Left hand side"color(white)(m)"Right hand side")$
$m C m m m m m m 16 m m m m m m m m 16$ $color(white)(m)"C"color(white)(mmmmmm)16color(white)(mmmmmmmm)16$
$m H m m m m m m 36 m m m m m m m m 36$ $color(white)(m)"H"color(white)(mmmmmm)36color(white)(mmmmmmmm)36$
$m O m m m m m m 50 m m m m m m m m 50$ $color(white)(m)"O"color(white)(mmmmmm)50color(white)(mmmmmmmm)50$

All atoms balance! The balanced equation is

$2 C_{8} H_{18} + 25 O_{2} \to 16 {CO}_{2} + 18 H_{2} O$ $2"C"_8"H"_18 +25"O"_2 → 16"CO"_2 + 18"H"_2"O"$

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Question #7ba58

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