How many mol Al2O3 made if: 4 Al (s) + 3 O2 (g) -> 2 Al2O3...

How many mol Al2O3 made if:

4 Al (s) + 3 O2 (g) -> 2 Al2O3 (s) ; 16 mol Al , 13 mol O2

* Al is limited reagent

Expert Solution

4 Al (s) + 3 O2 (g) --------------------> 2 Al2O3 (s)

4 3 2

16 13 ?

4 mol Al -------------> 3 mol O2

x mol Al --------------> 13 mol O2

x = 4 x 13 / 3 = 17.33 mol Al needed but we have only 16 mol Al . so limiting reagent is Al .

the product based on now limiting reagent

4 mol Al -------------------------> 2 mol Al2O3

16 mol Al ---------------> y mol Al2O3

y = 16 x 2 / 4 = 8 mol Al2O3

moles of Al2O3 made = 8

queen_honey_blossom answered 3 days ago

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