What volume of oxygen gas is needed to completely combust 0.202 L of butane gas?

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What volume of oxygen gas is needed to completely combust 0.202 L of butane gas?

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Answer 1 · 2017-02-16T06:28:24

We need a stoichiometric equation:

$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l)$ $C_4H_10(g) + 13/2O_2(g) rarr 4CO_2(g) + 5H_2O(l)$

Explanation:

The stoichiometrically balanced equation tells me unequivocally that for each equiv butane, I NEED $\frac{13}{2}$ $13/2$ equivs of dioxygen gas for complete combustion. Agreed?

So, since gaseous volume is proportional to the number of moles, at given temperature and pressure, we need a volume of dioxygen gas $= 0.202 \cdot L \times \frac{13}{2} = a bit under 1.5 litres$ $=0.202*Lxx13/2="a bit under 1.5 litres"$ . How many litres of carbon dioxide gas will result?

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What volume of oxygen gas is needed to completely combust 0.202 L of butane gas?

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