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

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 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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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 = = 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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The activity series is shown in the attached picture. if the element is below the other, then you can displace the other. following the activity series as a guide, the reactions are as follows: nickel and aluminum chloride3 ni + 2 alcl₃  → 3 nicl₂ + 2 albarium and copper bromideba + cubr  →  no reaction because cu is less oxidizable than bapotassium nitrate and aluminum3 kno₃ + al  → al(no₃)₃ + 3 ksodium phosphate and zincnaso₄ + zn  → znso₄ + na