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find the limiting reagent: *calculate the amount of aluminum oxide (in moles) that can be produced from the reaction of 12.0 g of aluminum with the excess of oxygen; *calculate the amount of aluminum oxide (in moles) that can be produced from the reaction of 11.0 g of oxygen with the excess of aluminum b) calculate the amount of aluminum oxide (in grams) that can be produced in this process c) calculate how many grams of excess reagent will be used in the process d) calculate how many grams of excess reagent will be left after the reaction

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rafikiedu08

The amount of aluminum oxide(in moles) produced by reaction of 12.0 g of aluminum is 0.458

MM (Al) = 26.98 g/mol  MM(O2) = 31.999 g/mol  MM(Al2O3) = 101.96 g/mol

The reaction is as follows;

4Al + 3O2 = 2Al2O3

Number of moles of Al = 12 / 26.98 = 0.445 mol

Number of moles of oxygen gas : 11 / 31.999 = 0.344 mol

3 mol oxygen gas needs 4 mol O2

So 0.344 mol O2 will need : ( 0.344 x 4) / 3 = 0.458 mol Al

So Al is the limiting reagent.

4 mol Al makes 2 mol Al2O3

So 0.445 mol Al will make : 0.445 / 2 = 0.222 mol Al2O3

Mass of Al2O3 in grams : mol x molar mass

= 0.222 x 101.96

= 22.7 grams

4 moles Al use 3 mol O2

So 0.445 mol Al will use : ( 0.445 x 3) / 4 = 0.333 mol O2 (excess reagent)

Mass of O2 used = 0.333 x 31.999 = 10.7 grams

Mass of excess O2 left after reaction = 11 - 10.7 = 0.3 grams

Yes this answer makes sense because the mass of total reactant species must be equal to the sum of product species and excess reagent.

So here : 12 +11 = 23 g reactant

product formed is 22.7 g and excess reagent is 0.3 grams , so total : 22.7 + 0.3 = 23 g

To learn more about limiting reagent visit:

brainly.com/question/11848702

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