P\(_4\)  + 50\(_2\) → P\(_4\)O\(_{10}\)

for the complete combustion of 124 grammes of phosphorus we need

5 x 22.4 litres of 0\(_2\)

for 12.4 gms of P\(_4\)  we need

\(\frac{5 \times 22.4}{124} \times \frac{12.4}{1}\)

\(\frac{22.4}{2}\) = 11.2 litres

Alternatively, 

  P\(_4\)  +  5 0\(_2\)  →  P\(_4\)O\(_{10}\)

  1 mole  : 5 moles

 Mass of P\(_4\)  = 12.4

Molar mass of P\(_4\)  = (31 x 4)=124g/mol

  n = \(\frac{mass}{Molar mass}\)

 = \(\frac{12.4}{124}\)

 = 0.1 mole

Then 0.1 mole P\(_4\)  will react with 0.5 mole 0\(_2\)

But  n = \(\frac{Volume}{Molar Volume}\)

  0.5 = \(\frac{Volume}{22.4}\)

 V = 0.5 X 22.4

  V = 11.20 Litres