Iron (III) Oxide reacts with Aluminum to form Aluminum Oxide and Iron. If I have 255 g of Aluminum and unlimited Iron (III) oxide, how many grams of Iron will form?
I am really not sure how to work with this. What does it mean when it says that the iron is unlimited, how would you calculate that in? I am unsure how to convert to moles and calculate grams on this one.
Fe2O3 + Al ---> Al2O3 + Fe
same number of mols of Fe as of Al
Al = 30 g/mol
so 225/30 mols of Al
so 225/30 mols of Fe
so
(225/30) mol mass of Fe
(225/30)56 = 420 grams of iron
To solve this question, we need to use stoichiometry to calculate the amount of iron that will form. Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction.
First, let's write the balanced chemical equation for the reaction:
Fe2O3 + 2Al -> 2Al2O3 + 2Fe
According to the equation, for every 1 mole of iron (III) oxide (Fe2O3), 2 moles of aluminum (Al), 2 moles of aluminum oxide (Al2O3), and 2 moles of iron (Fe) are produced.
To calculate the amount of iron that will form, we can use the following steps:
Step 1: Convert the given mass of aluminum (Al) to moles.
The molar mass of aluminum (Al) is 26.98 g/mol.
Number of moles of aluminum (Al) = 255 g / 26.98 g/mol.
Step 2: Use stoichiometry to determine the moles of iron (Fe) produced.
From the balanced equation, we know that for every 2 moles of aluminum (Al), 2 moles of iron (Fe) are produced.
Therefore, the number of moles of iron (Fe) = moles of aluminum (Al) x (2 moles of Fe / 2 moles of Al).
Step 3: Convert the moles of iron (Fe) to grams.
The molar mass of iron (Fe) is 55.85 g/mol.
Mass of iron (Fe) = moles of iron (Fe) x molar mass of iron (Fe).
By following these steps, you can calculate the grams of iron that will form in this reaction.