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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