Question

IRCICICLES TUTOR Limiting Reactants: Full Analysis For this reaction, 24.4 g sulfur dioxide reacts with 10.5 g oxygen gas. sulfur dioxide (g) + oxygen (g) — sulfur trioxide (g) What is the maximum mass of sulfur trioxide that can be formed? What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? Submit Show Approach Show Tutor Steps

Answer 1

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

Theoretical yield:

24.4g SO2x(1 mole SO2/64g so2)X 2molSO2/2 mol so2)X 80 g so3/1mol so3= 30.5 g SO3

10.5 g O2 X(1mol O2/32 g O2)(2 mol so3/1 mole O2)x80 g SO3/1 mol O so3= 52.5 g SO3

Hence SO2 forms less amount of SO3, limiting reagent= SO2

Amount of SO3: 30.5 g

moles of O2 consumed= 0.19 mol

Mass of O2 consumed= 0.19x32= 6.1 g

Mass of excess reagent(O2)= 10.5-6.1=4.4 g O2 left.