Question

Aluminum (AI) is reacted with oxygen to form aluminum oxide. If 21.066 g of aluminum reacts to form 39.626 g of aluminum oxide, determine the empirical formula for aluminum oxide.
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Answer 1

Answer:

Aluminum (AI) is reacted with oxygen to form aluminum oxide. If 21.066 g of aluminum reacts to form 39.626 g of aluminum oxide, determine the empirical formula for aluminum oxide.

Explanation:

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Answer 2

Empirical formula for Aluminum oxide is Al₂O₃

Explanation:

• Given Aluminum (AI) is reacted with oxygen to form aluminum oxide.

• If 21.066 g of aluminium reacts to form 39.626 g of aluminum oxide

• we have to find the empirical formula of aluminium oxide.
• A Given weight of aluminum is 21.066 gm
• The Molecular weight of Aluminium is 26.98 gm
• The number of moles of aluminum can be calculated as

• The Number of moles of Aluminium= Weight of Aluminium/molecular weight of Aluminium.

• Number of moles of Aluminium= 21.066/26.98
• Number of moles of Aluminium=0.780 moles

21.066 g of aluminum reacts to form 39.626 g of aluminum oxide so the amount of oxygen that  reacted to form aluminium oxide will be =39.626 -21.066 which will be 18.56.

• Given weight of oxygen is 18.56 gm
• The Molecular weight of oxygen is 16 gm
• Number of moles of Oxygen = Weight of oxygen/molecular weight of oxygen.
• Number of moles of oxygen=18.56/16

• Number of moles of oxygen=1.16 moles

0.780 moles of Aluminium reacts with 1.16 moles of Oxygen

we have to divide the number of moles of aluminium by number of moles of oxygen

=Aluminium/Oxygen

=0.780/1.16 which is equal to 0.672 that is 2/3

so the empirical formula Aluminium oxide is Al₂O₃