# Air contain 20 % O2 by volume. How many cm³ of air will be required for oxidation of 100 cm³ of acetylene?

Air contain 20 % $$\ce{O2}$$ by volume. How many cm³ of air will be required for oxidation of 100 cm³ of acetylene?

(a) 1064 cm³
(b) 212.8 cm³
(c) 500 cm³
(d) 1250 cm³

My solution:

equation: $$\ce{2C2H2 +5O2 -> 4CO2 +2H2O}$$

From stoichiometric calculations:

2 $$\ce{C2H2}$$ moles will react with 5 molecules of oxygen
1 $$\ce{C2H5 ->} 5/2$$ moles of $$\ce{O2}$$
$$22.4\ \mathrm{l}$$ $$\ce{C2H2 ->}$$ $$5/2\times22.4\ \mathrm{l}$$ of $$\ce{O2}$$

1 volume of air -> 0.2 (20/100) volume of oxygen

Now the only thing I don’t know is how to convert the volume into liters so that I can find the correct answer.

$$2\ \mathrm{mol}$$ $$\ce{C2H2}$$ will react with $$5\ \mathrm{mol}$$ of oxygen.
$$1\ \mathrm{mol}$$ $$\ce{C2H2}$$ will react with $$5/2\ \mathrm{mol}$$ of $$\ce{O2}$$.

If we suppose that the gases are under standard conditions, then
$$22.4\ \mathrm{L}$$ $$\ce{C2H2}$$ will react with $$\frac{5}{2} \times 22.4\ \mathrm{L }$$ $$\ce{O2}$$.

This means that: $$0.1\ \mathrm{L}$$ $$\ce{C2H2}$$ will react with $$\frac{5}{2} \times 0.1\ \mathrm{L}$$ $$ce{O2}$$.
$$0.1\ \mathrm{L}$$ $$\ce{C2H2}$$ will react with $$\frac{5}{2} \times 0.1\times 5\ \mathrm{L}$$ of air.
So, the volume of air is $$1.25\ \mathrm{L}= 1250\ \mathrm{cm^3}$$

What I suggest is we can use Gay Lussac's law of combining volumes of gases. According to this law, 2 volumes of acetylene reacts with 5 volumes of oxygen. From the balanced equation, the ratio between volumes of gaseous reactants and products are in the ratio 2:5:4.

2 volumes of $$\ce{C2H2}$$ requires 5 volumes of $$\ce{O2}$$.

The volume of $$\ce{O2}$$ used for $$\pu{100 cm3}$$ of $$\ce{C2H2} = (5/2) \times 100 = \pu{250 cm3}$$.

But air contains only 20% of $$\ce{O2}$$, i.e. $$0.2$$ of $$\ce{O2}$$.

Thus, the amount of air required to get $$\pu{250 cm3}$$ of oxygen $$= 250×100÷20 = \pu{1250 cm3}$$