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

Answer 2

Using the balanced equation, we can see that 1 mole of CO requires 1 mole of O2 for complete combustion. First, find the number of moles of CO from 277 g. Then, since the reaction occurs at STP, use the ideal gas law to convert moles of O2 to liters. The molar volume of a gas at STP is 22.4 L/mol.