Sulfur dioxide and molecular oxygen react to formSO 3 according to the following

Question

Sulfur dioxide and molecular oxygen react to formSO3 according to the following equation:
2SO2 + O2 <================>2SO3
In one experiment, 2.00 mol SO2 and 2.00mol O2 were initially present in a flask. If 1.76 molesof SO3 formed at equilibrium, the amount ofSO2 and O2 wouldbe?????
2SO2 + O2 <================>2SO3
In one experiment, 2.00 mol SO2 and 2.00mol O2 were initially present in a flask. If 1.76 molesof SO3 formed at equilibrium, the amount ofSO2 and O2 wouldbe?????
In one experiment, 2.00 mol SO2 and 2.00mol O2 were initially present in a flask. If 1.76 molesof SO3 formed at equilibrium, the amount ofSO2 and O2 wouldbe?????

Explanation / Answer

initialmoles             2         2                   0 change                  -2x      -x                   2x Equb.moles         2-2x       2-x                  2x Given the no . of moles of SO3 formed at Equilibrium is1.76 S0 2x = 1.76 moles       x = 0.88 moles The no . of moles of SO2 is = 2 - 2x = 2 - 2(0.88 ) =0.24 moles No . of moles ,n = mass / Molar mass So mass of SO2 ,m = 0.24 moles * 64g/mol   Molar mass of SO2 = 32 + 2 * 16 = 32 g/mol                                =15.36 g No . of moles of O2 is , n' = 2-x = 2-0.88 = 1.12 moles mass of O2 m' = no . of moles * Molar mass of O2                         = 1.12 mol * 32 g / mol                         = 35.84 g

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