next h.OZ Part 41 HW Core Chemistry Skill Cakculating Moles of Product from a Li

Question

next h.OZ Part 41 HW Core Chemistry Skill Cakculating Moles of Product from a Limting Reactant Core Chemistry Skill Calculating Moles of Product from a Limiting Reactant Part A When exposed to air, aluminum metal, Al reacts with oxygen, O2, to produce a protective coating of aluminum What is the theoretical yield of aluminum axide if 2 20 mol of aluminum metal is exposed to 1.95 mol of axygen? Express your answer with the appropriate units. Hints underneath. The balanced reaction is shown here alue Units Submit My Answers Give Up Part B This question will be shown after you complete previous question(s)

Explanation / Answer

Part A

4 Al + 3O2 -------> 2A2O3

stoichiometrically , 4 mole of Al react wih 3 mole of O2

therefore, 2.20 mole of Al react with 3/4*2.20 =1.65mole of O2

available O2 is 1.95

therefore, Al is limiting reagent and O2 is excess reagent

stoichiometrically, 4mole of Al give 2mole of Al2O3

2.20 mole of Al give 1.10mole of Al2O3

molar mass of Al2O3 = 101.96g/mol

mass of Al2O3 produced = 101.96g/mol * 1.10 =112.16g

therefore,

Theoretical yield = 112.16g

PART B

Answer

0.751 mole of P2O5 can be produced

Explanatin

P4 + 5O2 -------> 2P2O5

Stoichiometrically, 2mole of P2O5 is produced from 1mole of P4

Molar mass of P4 = 123.892g/mole

mass of P4 = 186g

no of mole of P4 = 186g/23.892g/mol=1.5013

1.5013 mole of P4 produce 1.5013/2= 0.75065 mole of P2O5

Part C

Answer

No of mole of P2O5 can be produced = 2.6 mol

Explanation

P4 + 5O2 ------> 2P2O5

stoichiometrically, 2mole of P2O5 produced from 5mole of O2

Mass of O2 = 208g

Molar mass of O2 = 31.998g/mol

No of mole of O2 produced = 208g/31.998g/mol = 6.500

6.500 mole of O2 produce (2/5)*6.500=2.6 mole of P2O5

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