Answer
2.29 atm
Work Step by Step
We find:
$n=35.8\,g\,O_{2}\times\frac{1\,mol\,O_{2}}{32.0\,g\,O_{2}}$
$=1.11875\,mol$
$V=12.8\,L$
$T=(46+273)\,K=319\,K$
$P=\frac{nRT}{V}=\frac{(1.11875\,mol)(0.08206\,L\,atm\,K^{-1}mol^{-1})(319\,K)}{12.8\,L}$
$=2.29\,atm$
