I need some gas law help in chemistry please!

the question states: Butane is used to fill gas tanks for heating. The following equation describes its combustion:
2 C4H10(g) + 13 O2(g) ---> 8 CO2(g) + 10 H2O(g)

and asks: If a tank contains 47.3 g of butane, what volume, in liters, of oxygen is needed to burn all the butane at 0.93 atm and 17C?

I set it up using the ideal gas law and got the volume of butane, but then what? :(

Molar mass butane = 58.12g/mol
47.3g butane = 47.3/58.12 = 0.8138 mol

This will react with 0.8138*13/2 = 5.2897 mol O2 required

Use gas equation to calculate volume
PV = nRT
P = pressure = 0.93
V = volume = ?
n = moles = 5.2897
R = 0.082057
T = 273+17 = 290K
Substitute
0.93V = 5.2897 *0.082057*290
V = 135.4 L O2 required

If you want to go the route using volume of butane:
P = 0.93
V = ?
n = 0.8138
R = 0.082057
T = 290

0.93V = 0.8138*0.082057*290
V = 20.82 L butane

That is where you got to.
From the equation , 2 volumes butane react with 13 volumes O2 ( for gases you can use volumes in the equation)
Therefore 20.82L butane react with 20.82*13.2 = 135.4 L O2 require - which is the same answer as above