Bob F. answered • 03/06/16
Tutor
4.6
(11)
A Highly Experienced Chemist and Instructor for a Tutor
Let's write the reaction first:
2 SO2(g) + O2(g) → 2SO3(g)
The limiting reactant here is oxygen - it is stated that the reaction is run with excess SO2
Looking at the mass of O2 consumed and the stoichiometric ratio, we can calculate the yield:
23.8 g O2 x ( 1 mole O2 / 32 g O2) x ( 2 mole SO3 / mole O2) x (80 g SO3 / mole SO3) = 119 g SO3
The reaction as stated yields 119 g SO3
