Answer:
5290.7 kg of Al₂O₃ are produced in the reaction of 2800 kg of Al
Explanation:
The reaction to produce alumina is:
4 Al + 3O₂ → 2 Al₂O₃
So, let's convert the mass of Al to moles (mass / molar mass)
2800 kg = 2800000 g (Then, 2.8 ×10⁶ g)
2.8 ×10⁶ g / 26.98 g/mol = 103780.5 moles of Al
Ratio is 4:2, so with the moles I have, I will produce the half of moles of alumina.
103780.5 mol / 2 = 51890.25 moles of Al₂O₃
Now we can convert these moles to mass ( molar mass . mol )
101.96 g/mol . 51890.25 mol = 5290729.8 g
5290729.8 g → 5290.7 kg
The mass of alumina Al₂O₃ formed from the reaction between Aluminum and oxygen is 5290.73 kg
The equation for the reaction that leads to the formation of alumina(Al2O3) can be expressed as:
[tex]\mathbf{2Al + \dfrac{3}{2}O_2 \to Al_2O_3}[/tex]
Given that:
the mass of aluminium Al = (2800 kg = 2800 × 1000) grams
= 2800000 grams
The molar mass of Aluminium Al = 26.98 g/mol
Number of moles of Al = 2800000 g/ 26.98 g/mol
Number of moles of Al = 103780.5782 moles
Since the ratio of the aluminium in the reactant and product is 2:1
There will be half moles of alumina;
∴
Moles of alumina will be:
[tex]\mathbf{\dfrac{1}{2} \times 103780.5782 \ moles}[/tex]
= 51890.2891 moles of Al₂O₃
Mass = number of moles × molar mass
The molar mass of Al₂O₃ = 101.96 g/mol
Mass of Al₂O₃= 51890.2891 g × 101.96 g/mol
Mass of Al₂O₃ = 5290733.877 grams
Mass of Al₂O₃ = (5290733.877/1000 ) kg
Mass of Al₂O₃ = 5290.73 kg
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