How would you balance the following equation: CH3CH2CH3(g)+O2(g) --> CO2(g) + H2O(g)?

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How would you balance the following equation: CH3CH2CH3(g)+O2(g) --> CO2(g) + H2O(g)?

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Answer 1 · 2015-12-04T00:49:37

We can rewrite this as:

$C_{3} H_{8} (g) + O_{2} (g) \to {CO}_{2} (g) + H_{2} O (g)$ $"C"_3"H"_8(g) + "O"_2(g) -> "CO"_2(g) + "H"_2"O"(g)$

I would initially start with the number of carbons because the number of carbons in ${CO}_{2}$ $"CO"_2$ is even, and the number of oxygens in $O_{2}$ $"O"_2$ is also even.

$C_{3} H_{8} (g) + O_{2} (g) \to 3 {CO}_{2} (g) + H_{2} O (g)$ $"C"_3"H"_8(g) + "O"_2(g) -> 3"CO"_2(g) + "H"_2"O"(g)$

Then, I would realize that there are $8$ $8$ hydrogens on the left, and I need $4 \times 2$ $4xx2$ on the right.

$C_{3} H_{8} (g) + O_{2} (g) \to 3 {CO}_{2} (g) + 4 H_{2} O (g)$ $"C"_3"H"_8(g) + "O"_2(g) -> 3"CO"_2(g) + 4"H"_2"O"(g)$

Lastly, I would balance the oxygen, because it is easy to only affect the number of oxygens. It only complicates things more to assign a number to $O_{2}$ $"O"_2$ first because you'll probably end up changing it again later.

$C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 {CO}_{2} (g) + 4 H_{2} O (g)$ $color(blue)("C"_3"H"_8(g) + 5"O"_2(g) -> 3"CO"_2(g) + 4"H"_2"O"(g))$

Last confirmation:

$C$ $"C"$ : $1 \times 3 = 3 \times 1$ $1xx3 = 3xx1$
$H$ $"H"$ : $1 \times 8 = 4 \times 2$ $1xx8 = 4xx2$
$O$ $"O"$ : $5 \times 2 = 3 \times 2 + 4 \times 1$ $5xx2 = 3xx2 + 4xx1$

Since this cannot be reduced any further, this is good to go.

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How would you balance the following equation: CH3CH2CH3(g)+O2(g) --> CO2(g) + H2O(g)?

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