Question

For the following reaction, identify the limiting reactant and calculate the grams of product produced when 40.0 grams of the first reactant and 25.0 grams of the second reactant are used: (4 pts)

2SO2 + O2 --> 2SO3

You will have two answers for this question –

Limiting reactant
amount of SO3 produced (grams)

Show work

Answer 1

Molar Mass of SO2 = 64.066

= 2×64.066 = 128.132

= 40/128.132= 0.312

Molar Mass of O2 = 31.998

25/31.998 = 0.781

Molar Mass of SO3= 80.06

= 80.06× 2 = 160.12

= 0.312 × 160.12= 49.96

= 0.781×160.12/2=62.55

= 112.51g

limiting reactant is SO₂