Question

11. How much aluminum can be produced from the \(1.00 \times 10^{3} \mathrm{~kg}\) mineral bauxite, which mostly \(\mathrm{Al}_{2} \mathrm{O}_{3} ?\)

Answer 1

Bauxite = Al2O3

The balanced reaction of production of Al from Al2O3

2Al2O3 = 4Al + 3O2

Mass of Al2O3 = 1.00*10^3 kg

Moles of Al2O3 = mass/molecular weight

= (1.00*10^3 kg) / (101.96 kg/kmol)

= 9.808 kmol

From the stoichiometry of the reaction

2 kmoles Al2O3 produces = 4 kmoles Al

9.808 kmol Al2O3 produces = 9.808*4/2 kmoles Al

= 19.616 kmol

Mass of Al = moles x molecular weight

= 19.616 kmol x 26.98 kg/kmol

= 529.23 kg

= 5.29 * 10^2 kg

Answer 2

Q11. Given : mass Al₂O₃ = 1.00 x 10³ kg

mass Al₂O₃ = 1.00 x 10³ kg * (1000 g / 1 kg)

mass Al₂O₃ = 1.00 x 10⁶ g

moles Al₂O₃ = (mass Al₂O₃) / (molar mass Al₂O₃)

moles Al₂O₃ = (1.00 x 10⁶ g) / (102 g/mol)

moles Al₂O₃ = 9803.9 mol

Al₂O₃ (s) $\rightarrow$ 2 Al (s) + 3/2 O₂ (g)

moles Al produced = 2 * (moles Al₂O₃)

moles Al produced = 2 * (9803.9 mol)

moles Al produced = 19607.8 mol

mass Al produced = (moles Al produced) * (molar mass Al)

mass Al produced = (19607.8 mol) * (27.0 g/mol)

mass Al produced = 529411.8 g

mass Al produced = 529411.8 g * (1 kg / 1000 g)

mass Al produced = 529 kg

Answer 3

The boonud lachen 15 2 A10,4 - HAL(9) + 302 Moloy mag of Alb - 10/96 g/mol solo aromic mass of AL -26.989/mol QY 101.969 of Aboz gives_4%26.989 of AL 1x 103 x 103 g of Al2O, gives ag of Al 1 000 000 X4 x 26.98 2x 101.96 529227114799 : : 529. 227 kg of Al is produad 5.292 X 103 kg of At.

Answer 4

moles of Al₂O3= mass molar mass of Al2O3 3 = 1X109 xlog - 9.81 X10% mol 101.96 gimos 10 Now, Imole Al2O3 contains a 2 mol Al 99.81X/0% mol Alyoz contains (22x9.81X109Jmol. All = 1960 mol Al Now, mam a moles y molar man of Al = 19620 mol X 27 glmol = 5.3x105g = 5.3x10²Kg] Al can be I produced