if 30.0 grams of molten iron (III) oxide reacts with 275 grams of aluminum, what is the mass of the iron produced.
Here is a solved example of a stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the mass of iron produced in the reaction between iron (III) oxide and aluminum, we need to first write and balance the chemical equation to understand the stoichiometry involved:
Fe2O3 + 2Al -> 2Fe + Al2O3
Now, we can use the molar masses and stoichiometry from the balanced equation to find the mass of iron produced.
1. Calculate the molar mass of Fe2O3 (Iron (III) oxide):
Molar mass of Fe = 55.85 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of Fe2O3 = (2 * 55.85) + (3 * 16.00) = 159.69 g/mol
2. Determine the number of moles of Fe2O3:
Moles of Fe2O3 = Mass of Fe2O3 / Molar mass of Fe2O3
= 30.0 g / 159.69 g/mol
≈ 0.188 moles
3. Use stoichiometry to find the moles of Fe produced:
According to the balanced equation, 1 mole of Fe2O3 produces 2 moles of Fe.
Moles of Fe = 2 * Moles of Fe2O3
= 2 * 0.188 moles
≈ 0.376 moles
4. Calculate the mass of Fe produced:
Mass of Fe = Moles of Fe * Molar mass of Fe
= 0.376 moles * 55.85 g/mol
≈ 21.04 g
Therefore, the mass of iron produced in this reaction is approximately 21.04 grams.