Question #88b25

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Question #88b25

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Answer 1 · 2015-05-15T04:05:21

The chemical formula for octane is $C_{8} H_{18}$ $"C"_8"H"_18"$ . A combustion reaction is one in which one reactant reacts with oxygen gas, $O_{2}$ $"O"_2"$ . Since octane is a hydrocarbon, the products will be carbon dioxide and hydrogen. The balanced equation for the combustion of octane gas is:

${2C}_{8} H_{18} (g) + {25O}_{2} (g)$ $"2C"_8"H"_18("g")+"25O"_2("g")$ $\to$ $rarr$ ${16CO}_{2} (g) + {18H}_{2} O(g)$ $"16CO"_2("g")+"18H"_2"O(g)"$

Explanation:

First place an 8 in front of ${CO}_{2}$ $"CO"_2$ because of the 8 carbon atoms in a molecule of octane, and a 9 in front of $H_{2} O$ $"H"_2"O"$ because of 18 H atoms divided by 2 atoms of O (in $O_{2}$ $"O"_2$ ).

$C_{8} H_{18} (g) + O_{2} (g)$ $"C"_8"H"_18("g")+"O"_2("g")$ $\to$ $rarr$ ${8CO}_{2} (g) + {9H}_{2} O(g)$ $"8CO"_2("g")+"9H"_2"O(g)"$

Balance the oxygen atoms on both sides of the equation. There are 25 O atoms on the right side of the equation. So we place a coefficient of $\frac{25}{2}$ $25/2$ in front of the $O_{2}$ $"O"_2$ molecule on the left.

$C_{8} H_{18} (g)$ $"C"_8"H"_18("g")$ $+$ $+$ $\frac{25}{2} O_{2} (g)$ $25/2"O"_2("g")"$ $\to$ $rarr$ ${8CO}_{2} (g) + {9H}_{2} O(g)$ $"8CO"_2("g")+"9H"_2"O(g)"$

There cannot be a fraction as a coefficient, so we need to multiply the equation times 2.

${2C}_{8} H_{18} (g) + {25O}_{2} (g)$ $"2C"_8"H"_18("g")+"25O"_2("g")$ $\to$ $rarr$ ${16CO}_{2} (g) + {18H}_{2} O(g)$ $"16CO"_2("g")+"18H"_2"O(g)"$

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Question #88b25

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