How many grams of oxygen are needed to completely react with 9.30 moles of aluminum: 4Al + 3O2 → 2Al2O3
Added by Shannon T.
Step 1
Given: Moles of aluminum (Al) = 9.30 moles Moles of O2 required per 4 moles of Al = 3 moles Moles of O2 = (9.30 moles Al) x (3 moles O2 / 4 moles Al) Moles of O2 = 6.975 moles Show more…
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