Can someone just check my work/answer please?
How many liters of oxygen would be required for the complete combustion of 1.0 gram of methane? Assume STP.
O2 + CH4 –> CO2 + 4H
Molar Mass of CH4= (12.01) + (1.01 * 4)= 16.05 mol CH4
1.0 g CH4 | 1 mol CH4 = 0.0623 mol CH4
| 16.05 g CH4
0.0623 mol CH4 | 1 mol O2 = 0.0623 mol O2
| 1 mol CH4
PV=nRT
1atm * V = 0.0623mol O2 * (0.0821L/atm * mol/K) * 273.15 K
1V = 1.39712
V = 1.4 L O2
thanks so much!!