Answer:
The answer to your question is: x = 0.185 mol of Al2O3
Explanation:
Data
10 g of Al and 19 grams of O2
Reaction: 4Al + 3O2 → 2Al2O3
4(27) 3(32) 2 ( 102)
108g 96g 204g
Limiting reactant
108g Al ------------------ 96g O2
10 g ----------------- x
x = (10 x 96) / 108 = 8.9 g of O2
Then limiting reactant is Al
So
108 g of Al ---------------------- 204 g of Al2O3
10 g ---------------------- x
x = (10 x 204) / 108 = 18.8 g of Al2O3
102 g of Al2O3 ----------------- 1 mol
18.8 ----------------- x
x = 0.185 mol of Al2O3
Considering the reaction stoichiometry and the definition of limiting reagent, the correct answer is option B): the total number of moles of Al₂O₃ produced is 0.185 moles.
The balanced reaction is:
6 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 each compound is:
So, by reaction stoichiometry, the following amount of mass of each compound participate in the reaction:
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 96 grams of O₂ reacts with 108 grams of Al, 19 grams of O₂ react with how much mass of Al?
[tex]mass of Al=\frac{19 grams of O_{2}x108 grams of Al }{96 grams of O_{2} }[/tex]
mass of Al= 21.375 grams
But 21.375 grams of Al are not available, 10 grams are available. Since you have less mass than you need to react with 19 grams of O₂, Al will be the limiting reagent.
Then you can apply the following rule of three to calculated the amount of moles of aluminum oxide produced, using the limitng reagent: if by stoichiometry 108 grams of Al produce 2 moles of Al₂O₃, 10 grams of Al will produce how many moles of Al₂O₃?
[tex]amount of moles of Al_{2} O_{3} =\frac{10 grams of Alx2moles of Al_{2} O_{3} }{108 grams of Al}[/tex]
amount of moles of Al₂O₃= 0.185 moles
Finally, the correct answer is option B): the total number of moles of Al₂O₃ produced is 0.185 moles.
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