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OpenStudy (anonymous):

how many moles of aluminum oxide can be produced from 12.8 moles of oxygen gas reacting with excess aluminum ? 4al (s) +3O2 >2AL2O3 (s)

OpenStudy (anonymous):

this should be in the chemistry group ;)

OpenStudy (anonymous):

You are given 12.8 moles of oxygen . Simply, just multiply: 12.8 mol O2(g) x ((2 mol Al2O3(s)) / 3 mol O2(g))=8.53 mol aluminum oxide.

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