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

Question

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

1 Answer

Impact of this question

2345 views around the world

You can reuse this answer
Creative Commons License

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 $water.$ $"water."$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question