Question #42736

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Question #42736

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Answer 1 · 2017-03-26T06:27:51

Let's assume that $100 mol$ $"100 mol"$ of hydrocarbon are combusted:

Explanation:

$30 C_{3} H_{8} + 70 C_{4} H_{10} + 605 O_{2} \to 370 C O_{2} + 470 H_{2} O$ $30C_3H_8+70C_4H_10 + 605O_2 rarr 370CO_2 + 470H_2O$

And of course I could divide this by 10 to get prime coefficients:

$3 C_{3} H_{8} + 7 C_{4} H_{10} + \frac{121}{2} O_{2} \to 37 C O_{2} + 47 H_{2} O$ $3C_3H_8+7C_4H_10 + (121)/2O_2 rarr 37CO_2 + 47H_2O$

Which is (?) balanced with respect to mass and charge. Is it?

So you need a tad over 6 equiv of dioxygen gas per 10 equiv hydrocarbon mix to combust the hydrocarbon completely. Anyway, come back for another serve if you are unsatisfied; you have not asked a question yet.

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Question #42736

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