When octane is completely combusted, what is the whole number ratio between the dioxygen reactant, and the product hydrogens?

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When octane is completely combusted, what is the whole number ratio between the dioxygen reactant, and the product hydrogens?

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Answer 1 · 2017-02-22T16:18:17

Well, clearly it's $25 : 18$ $25:18$ .

Explanation:

You have the balanced chemical equation:

$C_{8} H_{18} (l) + \frac{25}{2} O_{2} (g) \to 8 C O_{2} (g) + 9 H_{2} O (l)$ $C_8H_18(l) + 25/2O_2(g) rarr 8CO_2(g) + 9H_2O(l)$

And the stoichiometry precisely specifies that $25 equiv$ $"25 equiv"$ of oxygen gas are required for complete combustion of $2 equiv$ $"2 equiv"$ of $octane$ $"octane"$ , to give $16 equiv$ $"16 equiv"$ of $carbon dioxide$ $"carbon dioxide"$ and $18 equiv$ $"18 equiv"$ of $water.$ $"water."$

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When octane is completely combusted, what is the whole number ratio between the dioxygen reactant, and the product hydrogens?

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