Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) + 3O2 (g) →→ 2Al2O3 (s) The maximum amount of Al2O3 that can be produced from 2.5 g of Al and 2.5 g of O2 is ________ g.
Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide:
4Al (s) + 3O2 (g) 2Al2O3 (s)
The maximum amount of Al2O3 that can be produced from 2.5 g of Al and 2.5 g of O2 is ________ g.
A)5.3
B)7.4
C)4.7
D)9.4
E)5.0
7.4
Well, let's calculate it step by step. First, we need to determine the limiting reagent. To do that, we compare the moles of Al and O2 to the stoichiometric ratio in the balanced equation.
The molar mass of Al is 27 g/mol, so 2.5 g of Al is equal to 2.5 g / 27 g/mol = 0.093 mol.
The molar mass of O2 is 32 g/mol, so 2.5 g of O2 is equal to 2.5 g / 32 g/mol = 0.078 mol.
Now, let's compare the moles of Al and O2 to the stoichiometric ratio in the balanced equation. The stoichiometric ratio between Al and Al2O3 is 4:2, and the ratio between O2 and Al2O3 is 3:2.
For Al: 0.093 mol x (2 mol Al2O3 / 4 mol Al) = 0.0465 mol Al2O3.
For O2: 0.078 mol x (2 mol Al2O3 / 3 mol O2) = 0.052 mol Al2O3.
The limiting reagent is Al because it produces less Al2O3 compared to O2. So, the maximum amount of Al2O3 that can be produced is 0.0465 g.
Therefore, the answer is:
A) 5.3 g.
Remember, Al is the life of the reaction!
To find the maximum amount of Al2O3 that can be produced from 2.5 g of Al and 2.5 g of O2, we need to determine the limiting reactant.
First, we need to convert the given masses of Al and O2 to moles.
Molar mass of Al (Aluminum) = 26.98 g/mol
Molar mass of O2 (Oxygen) = 32.00 g/mol
Moles of Al = Mass of Al / Molar mass of Al
Moles of Al = 2.5 g / 26.98 g/mol
Moles of O2 = Mass of O2 / Molar mass of O2
Moles of O2 = 2.5 g / 32.00 g/mol
Next, we need to determine the stoichiometric ratio between Al and Al2O3 (based on the balanced equation).
4 moles of Al produce 2 moles of Al2O3.
Now, we can calculate the amount of Al2O3 that can be produced from the moles of Al.
Moles of Al2O3 = (Moles of Al / 4) * 2
Moles of Al2O3 = (2.5 g / 26.98 g/mol) / 4 * 2
Finally, with the moles of Al2O3, we can convert it to grams by multiplying it by the molar mass of Al2O3.
Mass of Al2O3 = Moles of Al2O3 * Molar mass of Al2O3
Now, we can substitute the values and calculate the maximum amount of Al2O3 that can be produced.
The correct answer is then provided as an option among A) 5.3 g, B) 7.4 g, C) 4.7 g, D) 9.4 g, and E) 5.0 g.
4 moles Al to 3 moles O2
molar masses ... Al - 27 g ... O2 - 32 g
moles Al ... 2.5 / 27 ... 0.093 ... dividing by 4 = .023
moles O2 ... 2.5 / 32 ... 0.078 ... dividing by 3 = .026
Al is the limiting reactant ... 0.069 moles of O2 are consumed ... 3 * .023
... 32 g * .069 = ? ... added to 2.5 g of Al