Answer: If you react 6.0g of aluminum with excess iron(III) oxide, you would form approximately 12.4g of iron.
Step-by-step explanation:
The reaction you're referring to is likely the thermite reaction, where aluminum reacts with iron(III) oxide to produce aluminum oxide and iron. The balanced chemical equation for this reaction is:
2Al + Fe2O3 → Al2O3 + 2Fe
From this equation, we can see that 2 moles of aluminum react with 1 mole of iron(III) oxide to produce 2 moles of iron.
First, we need to convert the mass of aluminum to moles. The molar mass of aluminum (Al) is approximately 26.98 g/mol.
Let's calculate the moles of Al and then use the stoichiometry of the reaction to find out the mass of iron produced.
The calculation shows that 6.0g of aluminum is approximately 0.222 moles.
From the balanced chemical equation, we know that 2 moles of aluminum produce 2 moles of iron. Therefore, 0.222 moles of aluminum will produce 0.222 moles of iron.
Now, let's convert the moles of iron to grams. The molar mass of iron (Fe) is approximately 55.85 g/mol. Let's do this calculation.
The calculation shows that 0.222 moles of iron is approximately 12.4 grams.
So, if you react 6.0g of aluminum with excess iron(III) oxide, you would form approximately 12.4g of iron.