To prepare 297.0g of chromium we need 154.21g of aluminum.
The preparation of chromium using aluminum powder is an exothermic process. It is the reaction where heat is released during the formation of product.
Therefore, the chemical reaction for the formation of chromium is:
Cr2O3(s) + 2Al(s) -> 2CrO(s) + Al2O3(s)
Here, we can see that 2 moles of aluminum produces 2 moles of chromium.
Hence,
Molar mass of aluminum = 27g
as 2 moles of aluminum is required then 2 x 27 = 54g.
Molar mass of chromium = 54g
As, 2 moles of chromium is obtained then 2 x 54 = 104g.
Therefore, by cross multiplying we get the amount of aluminum required,
mass of aluminum produced = 54 x 297.9/ 104
= 16038/104 = 154,21g.
Therefore, the required amount of aluminum is 154.21g
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