What mass of al2o3 (alumina, a finely divided white powder that is produced as billows of white smoke) is formed in the reaction of 2800 kg of aluminum?

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:

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:

= 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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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

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