The correct answer to the question is Option A. 44.6 g
The balanced equation for the reaction is given below:
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
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 = [tex]\frac{23.6 * 204}{108}[/tex] = 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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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:
The molar mass of the compounds is:
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
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?
[tex]mass of Al_{2} O_{3} =\frac{23.6 grams of Alx204 grams of Al_{2} O_{3}}{108 grams of Al}[/tex]
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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