Question

Using the equation, `Fe_2O_3 + 2Al -> 2Fe + Al_2O_3`, how many grams of `Al` are needed to completely react with 135 grams of `Fe_2O_3`?

Answer 1

`135g Fe_2 O_3 * (1 mol Fe_2 O_3) / (159.69g Fe_2 O_3) * (2 mol Al) / (1 mol Fe_2 O_3) * (26.98g Al) / (1 mol Al) = 45.6g Al`

Explanation:

The coefficients in the equation provide a molar ratio between the chemical species in the equation, so you must first convert the mass of iron (III) oxide to moles of iron (III) oxide by multiplying it by the inverse of its molar mass.

Then, by looking at the relationship between the moles of iron (III) oxide and aluminum, you can determine the number of moles of aluminum per mole of `Fe_2 O_3`.

Finally, convert the number of moles of aluminum to grams of aluminum by multiplying by its molar mass. After unit cancellation, you can determine that 45.6 g Al is necessary to completely react with 135 g `Fe_2 O_3`.