How much aluminum oxide and carbon are needed to prepare 617 grams of Al if the reaction proceeds to 81.2% yield? The balanced chemical equation is 2Al2O3 (s) + 3C(s)->4Al(s)+3CO2.

Answer 1

To find the amount of aluminum oxide needed, we first need to calculate the molar mass of Al2O3 and Al:

Al2O3:
Al: 26.98 g/mol
O: 16.00 g/mol x 3 = 48.00 g/mol
Total molar mass = 26.98 g/mol + 48.00 g/mol = 74.98 g/mol

Al:
26.98 g/mol

Now we can calculate the amount of Al2O3 needed to produce 617 grams of Al:

Moles of Al = Mass/Molar mass
Moles of Al = 617 g / 26.98 g/mol = 22.86 mol Al

From the balanced chemical equation, we can see that 2 moles of Al2O3 produce 4 moles of Al. This means that 1 mole of Al2O3 produces 4/2 = 2 moles of Al.

Moles of Al2O3 needed = Moles of Al / 2 = 22.86 mol / 2 = 11.43 mol

Now we can calculate the mass of Al2O3 needed:

Mass of Al2O3 = Moles of Al2O3 x Molar mass of Al2O3
Mass of Al2O3 = 11.43 mol x 74.98 g/mol = 857.39 g

Since the reaction proceeds to 81.2% yield, we need to adjust the amount of Al2O3 needed:

Adjusted mass of Al2O3 needed = Mass of Al2O3 / Yield
Adjusted mass of Al2O3 needed = 857.39 g / 0.812 = 1055.90 g

Therefore, 1055.90 grams of aluminum oxide are needed to produce 617 grams of aluminum.

Next, we can calculate the amount of carbon needed using the stoichiometry of the reaction:

From the balanced chemical equation, 3 moles of C produce 4 moles of Al. This means that 1 mole of C produces 4/3 moles of Al.

Moles of C needed = Moles of Al x (3/4)
Moles of C needed = 22.86 mol x (3/4) = 17.15 mol

Now we can calculate the mass of carbon needed:

Mass of C = Moles of C x Molar mass of C
Mass of C = 17.15 mol x 12.01 g/mol = 205.62 g

Adjusted mass of C needed = Mass of C / Yield
Adjusted mass of C needed = 205.62 g / 0.812 = 253.12 g

Therefore, 253.12 grams of carbon are needed to produce 617 grams of aluminum.