What volume of O2 gas (in L), measured at 782 mmHg and 35 ∘C, is required to completely react with 53.5 g of Al? Please help.?

4Al + 3O₂ → 2Al₂O₃
4 moles of Al reacts with 3 moles of O₂.

Molar mass of Al = 26.98 g/mol
No. of moles of Al = 53.5/26.98 mol
No. of moles of O₂ required = (53.5/26.98) × (3/4) = 1.486 mol

Consider the O₂ required :
No. of moles, n = 1.486 mol
Pressure, P = 782/760 atm
Absolute temperature, T = 273 + 35 = 308 K
Gas constant, R = 0.0821 atm L / (mol K)
Volume, V = ? L

PV = nRT
V = nRT/P

Volume of O₂, V = 1.486 × 0.0821 × 308 / (782/760) = 36.5 L

hmmm... two totally different answers so far. ok.. let's see which one (if either is correct).

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follow the process I outlined here
https://answers.yahoo.com/question/index?qid=20151020082535AAJS16h

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4 Al + 3 O2 ---> 2 Al2O3

then
.. 53.5g Al x (1 mol Al / 26.98g Al) x (3 mol O2 / 4 mol Al) = 1.487mol O2

then
.. PV = nRT
.. V = nRT/P
.. V = (1.487mol) x (0.08206 Latm/molK) x (35+273.15 K) / (782mmHg x 1atm/760mmHg)
.. V = 36.5L

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where did "simonizer" go wrong?

this
.. "Al + O2 ===> Al2O3"
doesn't look balanced to me

this
.. "moles O2 needed 1.98 moles x 2 = 3.97 moles O2 needed
.. (see balanced equation for mole ratio of 2:1)"
doesn't make sense given that non-balanced equation

this
.. "Volume of O2 needed is calculated from PV = nRT and V = nRT/V"
is wrong.. should be
.. V = nRT/P.... P not V in that denominator

this
.. "V = (3.97 mol)(62.36 L-mm/ºK-mol)(308ºK)/782 mm Hg"
is doubly incorrect should be
.. (63.36 L-mmHg / K-mol)(308K)
catch the diff?
.. mmHg
.. no ° sign on K's

Simonizer.... what happened?