4.35 L of Butane (C4H10) is combusted with 26.50 L of Oxygen gas. What volume of each product is formed by this reaction? What volume of excess reactant remains?

Md Ariful Islam

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2C4H10(g)+13O2(g)→8CO2(g)+10H2O(g)

Since the question specified that the combustion reaction of butane (C4H10), represented by the chemical equation in the question, took place at S.T.P. (standard temperature and pressure conditions), then we must know that at S.T.P. 1mol of any gas = 22.4dm^3 of that gas.

Now, the stoichiometry of the balanced chemical equation reveals that 2mol of butane reacts with 13mol of oxygen gas (O2).

At S.T.P., volume of butane that reacted = (2mol × 22.4dm^3) / 1mol

= 44.8dm^3 of butane

Also at S.T.P., the volume of oxygen gas that reacted = (13mol × 22.4dm^3) / 1mol

= 291.2dm^3 of oxygen gas

Therefore, we have that at S.T.P., 44.8dm^3 of butane will react with 291.2dm^3 of oxygen for a complete combustion.

We were given 10cm^3 of butane which equals 0.01dm^3 of butane.

(N.B.: 1000cm^3 = 1dm^3)

If 44.8dm^3 of butane requires 291.2dm^3 of oxygen; therefore 0.01dm^3 of butane will require (0.01 × 291.2) / 44.8 dm^3 of oxygen

= (2.912 / 44.8) dm^3 of oxygen

= 0.065dm^3 (or 65cm^3 when converted) of oxygen

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