Question

Consider the following reaction: 2SO2(g)+O2(g)→2SO3(g) 290.2 mL of SO2 is allowed to react with 151.8 mL of O2 (both measured at 318 K and 47.9 mmHg ), what is the limiting reactant? What is the theoretical yield of SO3? If 170.7 mL of SO3 is collected (measured at 318 K and 47.9 mmHg ), what is the percent yield for the reaction?

Answer 1

Limiting reactant is SO2

Theoretical yield of SO3 = 0.0007006mol

percent yield of SO3 = 58.82%

Explanation

Ideal gas equation is

PV = nRT

number of moles, n = PV/RT

P = 47.9mmHg = 0.0630

T = 318K

R = 0.082057L atm/mol K

number of moles of SO2 = ( 0.0630atm × 0.2902L/0.082057(L atm/mol K) × 318K)

number of moles of SO2 = 0.0007006 mol

Number of moles of O2 = (0.0630atm ×0.1518L/0.082057( L atm /mol K ) × 318K ) = 0.0003665 mol

2SO2(g) + O2(g) -----> 2SO3(g)

stoichiometrically, 2moles of SO2 react with 1mole of O2

Therefore,

0.0003665 moles of O2 require 2×0.0003665 = 0.000733 moles of SO2 but available moles of SO2 is 0.0007006 only

Therefore,

SO2 is limiting reagent

stoichiomertically 2 moles of SO3 are obtained from 2moles of SO2

Therefore

0.0007006 moles of SO2 gives 0.0007006 moles of SO3

Theoretical yield = 0.0007006 mol

No of moles of SO3 actually obtained =( 0.0630atm × 0.1707L/(0.082057(L atm/mol K)× 318K) = 0.0004121 mol

percent yield = (Actual yield /Theoretical yield )×100

percent yield = (0.0004121mol/0.0007006mol)×100 = 58.82%