How many liters of oxygen are necessary for the combustion of 277 g of carbon monoxide assuming that the reaction occurs at stp the balanced equation is 2 co plus o2 equals 2 co2?

Answer 1

2CO + O2 -> 2CO2

find moles O2 gas

277 grams carbon monoxide (1 mole CO/28.01 grams)(1 mole O2/2 mole CO)

4.9447 moles O2

Now, since I am lazy, I use PV = nRT

V = nRT/P

V = (4.9447 moles O2)(0.08206 L*atm/mol*K)(298.15 K)/(1 atm)

V = 120.978 Liters of oxygen, or, in sigi figi;

121 Liters of oxygen needed for this combustion