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

Question

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

A) How many O2 molecules are needed to react with 3.25 g of S? B) What is the theoretical yield of SO3 produced by the quantities described in Part A?

2) 1.05 g H2 is allowed to react with 9.71 g N2, producing 1.15 g NH3.

A) What is the theoretical yield in grams for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units.

B) What is the percent yield for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units.

Thank you so much in advance!

Explanation / Answer

1) 2 moles of S reacts with 3 moles of O2

moles of S = 3.25 g/32.065 g/mol = 0.10 mols

moles of O2 needed = 0.10 x 3/2 = 0.15 mols

A) molecules of O2 required = 0.15 x 6.022 x 10^23 = 9.16 x 10^22 molecules

B) Theoretical yield of SO3 = 0.10 x 80.066 g/mol = 8.00 g

2) N2 + 3H2 ---> 2NH3

A) moles of N2 = 9.71 g/28 g/mol = 0.35 mols

moles of H2 = 1.05 g/2.016 g/mol = 0.52 mols

For complete consumption of N2 we ould need = 0.35 x 3 = 1.05 mols of H2

For complete consumption of H2 we would need = 0.52/3 = 0.17 mols of N2

Since moles of H2 is less than needed, this is limiting reagent

Theoretical yield for NH3 = 0.52 x 2 x 17/3 = 5.893 g

B) percent yield for the reaction = 1.15 x 100/5.893 = 19.515%

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