(SHOW CALCULATIONS) Aluminum oxide forms when aluminum reacts with oxygen by the following balanced reaction: 4 Al(s) + 3 O₂(g) →→→ 2 Al₂O3(s) What is the percent yield if 87.3 g of Al2O3 (MM = 101.96 g/mol) is recovered from reacting 55.0 g of Al (MM 26.98 g/mol) with an excess of O2?

(SHOW CALCULATIONS) Aluminum oxide forms when aluminum reacts with oxygen by the following balanced reaction: 4 Al(s) + 3 O₂(g) →→→ 2 Al₂O3(s) What is the percent yield if 87.3 g of Al2O3 (MM = 101.96 g/mol) is recovered from reacting 55.0 g of Al (MM 26.98 g/mol) with an excess of O2?

Chemistry for Engineering Students

4th Edition

ISBN:9781337398909

Author:Lawrence S. Brown, Tom Holme

Publisher:Lawrence S. Brown, Tom Holme

Chapter4: Stoichiometry

Section: Chapter Questions

Problem 4.87PAE

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Concept explainers

Stoichiometry

Stoichiometry is a chemical division that evaluates the quantitative content of the chemical process with the help of the reacting molecules or products taking part in the reaction. The term “stoicheion” in “Stoichiometry”, is an old Greek term for "element" whereas “metron” means "measure".

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(SHOW CALCULATIONS)
Aluminum oxide forms when aluminum reacts with oxygen by the following balanced reaction:
4 Al(s) + 3 O₂(g) →→→ 2 Al₂O3(s)
What is the percent yield if 87.3 g of Al2O3 (MM = 101.96 g/mol) is recovered from reacting 55.0 g
of Al (MM 26.98 g/mol) with an excess of O2?

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