Answer to Question #215204 in General Chemistry for Liana

Question #215204
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)

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1
Expert's answer
2021-07-09T05:15:08-0400

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

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