Elemental S reacts with O2 to form SO3 according to the reaction 2S+3O22SO3 A) H

Question

Elemental S reacts with O2 to form SO3 according to the reaction

2S+3O22SO3

A) How many O2 molecules are needed to react with 4.13 g of S?

Express your answer numerically in units of molecules.

B) What is the theoretical yield of SO3 produced by the quantities described in Part A?

Express your answer numerically in grams.

C) Next, consider a situation in which all of the S is consumed before all of the O2 reacts, or one in which you have excess S because all of the O2 has been used up.

For each of the given situations, indicate whether S or O2 is the limiting reactant.

- 3.00 mol Sulfur, 3.00 mol Oxygen

-3.00 mol Sulfur, 4.00 mol Oxygen

-3.00 mol Sulfur, 5.00 mol Oxygen

Explanation / Answer

                                   2S                      +                          3O2                                           2SO3

General equation 2*32 = 64g                                 3*32 = 96g                                               2*80 = 160g

in this problem           4.13g                            (4.13*96)/64= 6.195g

Part A) 64g of sulfur require 96 g of O2 to produce 160g of SO3

4.13g of sulfur requires (4.13*96)/64 = 6.195g of O2

Moles = weight /molecular weight = 6.195/32 = 0.1936mole of O2

one mole = 6.023x1023 molecules

Part B) 64g of Sulfur can produce 160g of SO3

            4.13g of sulfur produce (4.13*160)/64 = 10.325 g of SO3

0.1936 mole = 0.1936 * 6.023x1023 molecules = 1.166x1023 molecules

Part C) 2 mole S will completly react with 3 moles of O2

So 3 mole S require 4.5 moles of Oxygen

i) 3.00 mol Sulfur, 3.00 mol Oxygen ( limiting reagent is Oxygen )

ii) 3.00 mol Sulfur, 4.00 mol Oxygen ( limiting reagent is Oxygen )

iii) 3.00 mol Sulfur, 5.00 mol Oxygen ( limiting reagent is Sulfur )

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