If 31.5 of molten iron(III) oxide reacts with 175 of aluminum, what is the mass of iron produced?
Write and balance the equation.
Convert 175 g Al to moles. moles - grams/molar mass
Convert 31.5 g iron(III) oxide to oles.
Using the coefficients in the balanced equation, convert moles Al to moles iron.
Do the same for ferric oxide.
You will obtain two answers; of course only one can be correct. The correct answer in limiting reagent problems is ALWAYS the smaller one and the reagent producing that number is the limiting reagent. The convert the smaller value to grams. g = moles x molar mass.
To solve this problem, we need to use the balanced chemical equation for the reaction between iron(III) oxide (Fe2O3) and aluminum (Al). The balanced equation is:
2Al + Fe2O3 -> Al2O3 + 2Fe
From the balanced equation, we can determine that the molar ratio between Fe2O3 and Al is 1:2. This means that for every 1 mole of Fe2O3, we need 2 moles of Al.
First, we can calculate the number of moles of Fe2O3 and Al given the given masses:
Mass of Fe2O3 = 31.5 g
Molar mass of Fe2O3 = 2 * (55.85 g/mol) + 3 * (16 g/mol) = 159.7 g/mol
Moles of Fe2O3 = Mass / Molar mass = 31.5 g / 159.7 g/mol
Mass of Al = 175 g
Molar mass of Al = 26.98 g/mol
Moles of Al = Mass / Molar mass = 175 g / 26.98 g/mol
Now, we compare the moles of Fe2O3 and Al using the mole ratio:
Moles of Fe2O3 / Moles of Al = (31.5 g / 159.7 g/mol) / (175 g / 26.98 g/mol)
Simplifying, we get:
Moles of Fe2O3 / Moles of Al = 0.196
Since the molar ratio between Fe2O3 and Fe is also 1:2, this means that for every 1 mole of Fe2O3, we produce 2 moles of Fe.
Therefore, the moles of Fe produced will be twice the moles of Fe2O3:
Moles of Fe = Moles of Fe2O3 * 2 = 0.196 * 2
To calculate the mass of Fe produced, we multiply the moles of Fe by its molar mass:
Mass of Fe = Moles of Fe * Molar mass of Fe
Now you can substitute the values and calculate the mass of Fe produced.