Consider the reaction: 2SO2(g)+O2(g)→2SO3(g). If 288.7 mL of SO2 reacts with 150.7 mL of O2 (both measured at 312 K and 77 mbar) What is the theoretical yield of SO3?
Volume of SO2 = 288.7 mL = 0.2887L
VOlume of O2 = 150.7 mL = 0.1507 L
Temperature (T) = 312K
Pressure (P) = 77 mbar =0.07599 atm
Gas constant (R) = 0.0820 L.atm/mol.K
Ideal gas equation is
PV = nRT
Moles of SO2 are
n = 0.07599 atm * 0.2887L / ( 0.0820 L.atm/mol.K * 312K) = 8.57*10-4 mol
Moles of O2
n = 0.07599 atm * 0.1507L / ( 0.0820 L.atm/mol.K * 312K) = 4.47 *10-4 mol
NOw, reaction is :
2 SO2 + O2
2 SO3
2 mol 1 mol 2 mol
Thus, 8.57*10-4 mol of SO2 reacts with 8.57*10-4 mol/2 = 4.28 * 10-4 mol of O2
So, SO2 is a limiting reagent.
From reaction
2 mol of SO2 gives 2 mol of SO3
Thus, 8.57*10-4 mol of SO2 gives 8.57*10-4 mol of SO3
thus, theoretical yield of SO3 is 8.57*10-4 mol
Get Answers For Free
Most questions answered within 1 hours.