Iron can be obtained by reducing iron (3) oxide with aluminum.

Iron can be obtained by reducing iron (3) oxide with aluminum. Calculate the mass of aluminum and iron oxide (3) required to obtain iron weighing 140 g

Let us write the reaction equation for the interaction of aluminum and iron oxide (3):
Fe2O3 + 2Al = Al2O3 + 2Fe
According to the law of conservation of matter by reaction, we get:
ν (Al) = ν (Fe)
ν (Al) = 2ν (Fe2O3)
ν = m / M
m (Al) / M (Al) = m (Fe) / M (Fe)
m (Al) / (2 * M (Al)) = m (Fe2O3) / M (Fe2O3)
Let’s define the molar masses:
M (Al) = 26 g / mol.
M (Fe) = 56 g / mol
M (Fe2O3) = 56 * 2 + 16 * 3 = 160 g / mol
Determine the mass of aluminum:
m (Al) = m (Fe) * M (Al) / M (Fe) = 140 * 26/56 = 65 g.
Determine the mass of iron oxide (3):
m (Fe2O3) = m (Al) * M (Fe2O3) / (2 * M (Al)) = 65 * 160 / (2 * 26) = 200 g.
Answer: you need 65 g of Al and 200 g of Fe2O3.




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