Air contains 20 % O2 by volume. How many cm³ of air will be required for oxidation of 100 cm³ of acetylene?
Let's begin with your 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 $1.25\ \mathrm{L}= 1250\ \mathrm{cm^3}$