First, let's write the combustion reaction of ethane equation:
2 C2H6 + 7 O2 --> 4 CO2 + 6 H2O
Now we need to transform 15 cm³ into liters.
1 cm³ ---- 0.001 liter
15 cm³ ---- x liter
x = 0.015 liters
We know that 1 mol of any gas in STP occupies the volume of 22.4L. So let's transform liters of ethane into mol.
1 mol --- 22.4 L
x mol --- 0.015 L
x = 6.696 x 10^-4 mol of C2H6
Now we use the equation ratio to find out the quantity of oxygen in moles:
2 mole C2H6 --- 7 mole O2
6.696 x 10^-4 mol of C2H6 --- x mol of O2
x = 2.34 x 10^-3 mol of O2
Now let's transform into liters:
2.34 x 10^-3 mol of O2 --- x liters
1 mol --- 22.4 L
x = 0.0525 liters or 52.5 cm³
Answer: The volume of oxygen required is 52.5 cm³.