Answer
2.11 mol
Work Step by Step
$R=0.08206\,L\,atm\,mol^{-1}K^{-1}$
$P=10.5\,atm$
$V=5.00\,L$
$T=(30.0+273)\,K=303\,K$
From the ideal gas law, we have
$ n=\frac{PV}{RT}=\frac{10.5\,atm\times5.00\,L}{0.08206\,L\,atm\,mol^{-1}K^{-1}\times303\,K}$
$=2.11\,mol$
