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Chromium metal can be produced from the high temperature reaction of Cr2O3 with silicon by the following reaction: Cr2O3(s) + 2Al(l) --> 2Cr(l) + Al2O3(s) Calculate the number of grams of aluminum required to prepare 500.0g of chromium metal

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MissPhiladelphia
To solve this problem, we need to know the molar mass of Aluminum and Chromium.

Aluminum = 27 g/mol
Chromium = 52 g/mol

Then, we do stoichiometry:

500.0 g Cr x (1/52 g/mol Cr) x (2 mol Al/2 mol Cr) x (27 g/mol Al)

= 259.62 g aluminum

Thus, you need 259.62 grams of reactant aluminum.

The number of grams of aluminium required to prepare 500.0g of chromium metal is 259.62 g

Balanced chemical reaction

Cr₂O₃(s) + 2Al(l) --> 2Cr(l) + Al₂O₃(s)

According to the chemical reaction, 2 moles of aluminium produces 2 moles of chromium.

molar mass of aluminium = 27

Therefore,

2 moles of aluminium = 2 × 27 = 54 g

molar mass of chromium = 52 g

2 moles of chromium = 2 × 52 = 104 g

Therefore,

104 g of Cr requires 54 g of Al

500 g of Cr will need ?

cross multiply

mass of aluminium produced = 54 × 500 / 104

mass of aluminium produced =27000 / 104

mass of aluminium produced = 259.615384615

mass of aluminium produced = 259.62 g

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