Answer
676 g
Work Step by Step
We find:
$PV=nRT$ (Ideal gas law)
$\implies n=\frac{PV}{RT}=\frac{(7.50\,atm)(50.0\,L)}{(0.0820575\,L\cdot atm\cdot mol^{-1}\cdot K^{-1})(25+273.15)K}=15.32774\,mol$
Grams of propane= $15.32774\,mol\,C_{3}H_{8}\times\frac{44.1\,g}{1\,mol\,C_{3}H_{8}}=676\,g$
