chemistry stoichiometry

Question

In order to produce 495 g of water, how many liters of oxygen gas are required if the oxygen is at 116.57 kPa and 32 degrees Celsius?

Answer 1

First, find moles of O₂ needed from 2 H₂ + O₂ --> 2 H₂O: (495g H₂O) * (1 mol H₂O /18.02g H₂O) * (1 mol O₂ /2 mol H₂O) = 13.73 mol O₂.

From the ideal gas law PV = nRT, V = nRT/P = (13.73 mol) (8.3144 L kPa/mol K) (32 + 273.15 K) / (116.57 kPa) = 299 L.