Chemistry, 20.04.2020 14:30 jflandersneongr

Aluminum Oxide is formed when aluminum combines with oxygen in the air. How many grams of Al2O3 are formed when 23.6g of Al reacts completely with oxygen?
4 Al + 3 O2 —> 2 Al2O3
A. 44.6 grams B. 35.6 grams C. 87.6 grams D. 21.6 grams

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Answer from: annabanana1298

Taking into account the reaction stoichiometry, the correct answer is option A. 44.6 grams.

In first place, the balanced reaction is:

4 Al + 3 O₂ → 2 Al₂O₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Al: 4 moles O₂: 3 moles Al₂O₃: 2 moles

The molar mass of the compounds is:

Al: 27 g/mole O₂: 32 g/mole Al₂O₃: 102 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Al: 4 moles× 27 g/mole= 108 gramsO₂: 3 moles× 32 g/mole= 96 grams Al₂O₃: 2 moles× 102 g/mole= 204 grams

Then the following rule of three can be applied: if by reaction stoichiometry 108 grams of Al form 204 grams of Al₂O₃, 23.6 grams of Al form how much mass of Al₂O₃?

$mass of Al_{2} O_{3} =\frac{23.6 grams of Alx204 grams of Al_{2} O_{3}}{108 grams of Al}$

mass of Al₂O₃= 44.6 grams

Then, 44.6 grams of Al₂O₃ are formed when 23.6 g of Al reacts completely with oxygen.

Finally, the correct answer is option A. 44.6 grams.

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Answer from: Bayabbay6097

The correct answer to the question is Option A. 44.6 g

The balanced equation for the reaction is given below:

4Al + 3O₂ —> 2Al₂O₃

Next, we shall determine the mass of Al that reacted and the mass of Al₂O₃ produced from the balanced equation.

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 4 × 27 = 108 g

Molar mass of Al₂O₃ = (27×2) + (3×16)

= 54 + 48

= 102 g/mol

Mass of Al₂O₃ from the balanced equation = 2 × 102 = 204 g

SUMMARY:

From the balanced equation above,

108 g of Al reacted to produce 204 g of Al₂O₃.

Finally, we shall determine the mass of Al₂O₃ produced by the reaction of 23.6 g of Al. This can be obtained as follow:

From the balanced equation above,

108 g of Al reacted to produce 204 g of Al₂O₃.

Therefore,

23.6 g of Al will react to produce = $\frac{23.6 * 204}{108}$ = 44.6 g of Al₂O₃

Thus, 44.6 g of Al₂O₃ were obtained from the reaction.

Hence, Option A. 44.6 g gives the correct answer to the question.

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Aluminum Oxide is formed when aluminum combines with oxygen in the air. How many grams of Al2O3 are...

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