A student combusted 0.500g of purified aluminum power with excess oxygen in an oxygen atmosphere according to the reaction, 4Al(s) + 3O2(g)= 2Al2O3(s)

A) What is the limiting reactant?
B) How many moles of Al were used?
C) How many moles of Al2O3 would form,
based on the moles of the limiting
reactant?
D) How many moles of O2 are required to
react completely with the aluminum?
E) How many grams of O2 are required to
react completely with the aluminum?
F) What is the theoretical yield of
AL2O3 in grams?
G) If the student collected 0.918 grams
of AL2O3 product, what was the
present yield of Al2O3 obtained?

See your PbI2 problem above.

A) To determine the limiting reactant, we need to compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.

First, let's convert the mass of Al to moles:
0.500g Al * (1 mole Al / molar mass Al) = X moles Al

Next, let's convert the mass of O2 to moles:
Since there is an excess of O2, we can assume it is not the limiting reactant.

B) To calculate the moles of Al used, we need to know the molar mass of Al. The molar mass of Al is 26.98 g/mol.

X moles Al * (molar mass Al / 1 mole Al) = Y moles Al

C) From the balanced equation, we can see that the molar ratio between Al and Al2O3 is 4:2, or 2:1.
Thus, the number of moles of Al2O3 formed should be half the number of moles of Al used:
Y moles Al * (1 mole Al2O3 / 2 moles Al) = Z moles Al2O3

D) The balanced equation tells us that the molar ratio between Al and O2 is 4:3.
Thus, the number of moles of O2 required to react with the aluminum can be calculated as follows:
Y moles Al * (3 moles O2 / 4 moles Al) = W moles O2

E) To calculate the mass of O2 required to react completely with aluminum, we need to know the molar mass of O2. The molar mass of O2 is 32.00 g/mol.

W moles O2 * (molar mass O2 / 1 mole O2) = V grams O2

F) The theoretical yield of Al2O3 can be calculated by multiplying the moles of Al2O3 formed by its molar mass. The molar mass of Al2O3 is 101.96 g/mol.

Z moles Al2O3 * (molar mass Al2O3 / 1 mole Al2O3) = U grams Al2O3

G) The present yield of Al2O3 obtained can be calculated by dividing the actual mass of Al2O3 collected by the theoretical yield, and then multiplying by 100%:

(actual yield / theoretical yield) * 100% = Present yield of Al2O3

Let's calculate each step one by one:

A) The limiting reactant is the reactant that is completely consumed and limits the amount of product formed. We can compare the moles of Al and O2 to determine the limiting reactant.

To find the moles of Al:
0.500g Al * (1 mol Al / 26.98 g Al) = 0.0185 moles Al

Since the balanced equation ratio is 4:3, this means that 0.0185 moles of Al would require (0.0185 moles Al * 3/4) = 0.0139 moles of O2 for complete reaction.

Next, we convert the moles of Al2O3 formed by multiplying the moles of Al by the stoichiometric ratio:
0.0185 moles Al * (1 mol Al2O3 / 4 mol Al) = 0.00463 moles Al2O3.

Therefore, the limiting reactant is Al, as it produces less moles of Al2O3 compared to the stoichiometric ratio.

B) The number of moles of Al used is 0.0185 moles.

C) Based on the moles of the limiting reactant (Al), the number of moles of Al2O3 formed would be half that amount. Therefore, 0.0185 moles of Al would form 0.00925 moles of Al2O3.

D) The balanced equation states that 4 moles of Al react with 3 moles of O2. Since we know that 0.0185 moles of Al were used, we can calculate the required moles of O2:
0.0185 moles Al * (3 moles O2 / 4 moles Al) = 0.0139 moles O2.

E) To calculate the grams of O2 needed, we multiply the moles of O2 by its molar mass:
0.0139 moles O2 * (32.00 g O2 / 1 mole O2) = 0.4448 grams O2.

F) The theoretical yield of Al2O3 can be calculated by multiplying the moles of Al2O3 formed (0.00925 moles) by its molar mass:
0.00925 moles Al2O3 * (101.96 g Al2O3 / 1 mole Al2O3) = 0.942 grams Al2O3.

G) The present yield of Al2O3 obtained can be calculated by dividing the actual mass of Al2O3 collected (0.918 grams) by the theoretical yield (0.942 grams), and then multiplying by 100%:
(present yield / theoretical yield) * 100% = (0.918 g / 0.942 g) * 100% = 97.5% (approximately).

Therefore, the present yield of Al2O3 obtained is 97.5%.

A) To determine the limiting reactant, we need to compare the moles of Al and O2. First, we convert the mass of Al to moles:

- The molar mass of Al is 26.98 g/mol.
- Therefore, moles of Al = (0.500 g Al) / (26.98 g/mol Al).
- Calculate this value to find the moles of Al.

Next, we convert the moles of O2 to moles:
- Based on the balanced equation, the stoichiometry of the reaction is 4:3 (four moles of Al react with three moles of O2).
- Therefore, moles of O2 = (moles of Al) * (3 moles O2 / 4 moles Al).
- Calculate this value to find the moles of O2.

Finally, compare the moles of Al and O2. The reactant with the lesser number of moles is the limiting reactant.

B) To determine the number of moles of Al used, use the value we calculated in step A.

C) Based on the stoichiometry of the reaction, the molar ratio of Al2O3 to Al is 2:4 (2 moles of Al2O3 form from 4 moles of Al). Therefore, multiply the moles of Al by the ratio to obtain the moles of Al2O3 that would form.

D) To calculate the moles of O2 required to react completely with the aluminum, use the stoichiometry of the balanced equation. The ratio of moles of O2 to Al is 3:4 (3 moles of O2 react with 4 moles of Al). Multiply the moles of Al by the ratio to obtain the moles of O2 required.

E) To convert the moles of O2 required into grams, use the molar mass of O2. The molar mass of O2 is 32.00 g/mol. Multiply the moles of O2 by the molar mass to obtain the grams of O2 required.

F) To calculate the theoretical yield of Al2O3 in grams, use the stoichiometry of the balanced equation. The ratio of moles of Al2O3 to Al is 2:4 (2 moles of Al2O3 form from 4 moles of Al). Multiply the moles of Al by the molar mass of Al2O3 to obtain the theoretical yield of Al2O3 in grams.

G) To calculate the percent yield of Al2O3 obtained, divide the actual yield (0.918 grams) by the theoretical yield (calculated in question F) and multiply by 100.

4al+3o2------------->2al2o3

Mass of al=2.7g
Moles of al=2.7÷27=0.1 moles
Comparing al and o2. According to balance equation
Al : o2
4 : 3
1 : 3÷4
0.1 moles : 0.75×0.1 moles
0.1 moles : 0.075 moles
So.
Mass of o2=moles×molar mass
=0.075×32
Mass=2.4g