Final answer:
69.9 g of iron is needed to form 100.0 g of rust. To find the mass of iron needed to form 100.0 g of rust, we use stoichiometry to convert the mass of Fe2O3 to moles, apply the mole ratio from the balanced equation, and then convert moles of Fe back to mass, yielding 69.9 g of iron.
Step-by-step explanation:
To determine the mass of iron needed to produce 100.0 g of rust, we will use stoichiometry and known molar masses.
Step-by-step Calculation
First, we must recognize that rust is iron(III) oxide hydrate (Fe2O3·xH2O).
For simplicity, we will consider the anhydrous form Fe2O3.
The molar mass of Fe2O3 is 159.7 g/mol.
Second, we use the mole ratio from the balanced chemical equation.
We know that 4 moles of iron (Fe) are required to produce 2 moles of Fe2O3.
To find the mass of iron required, we first convert 100.0 g of Fe2O3 to moles and then use the stoichiometric mole ratio to find the moles of Fe needed.
Finally, we multiply the moles of Fe by its molar mass which is 55.85 g/mol to find the mass of Fe required.
Example Calculations
Using the molar mass of Fe2O3, convert 100.0 g rust to moles:
100.0 g Fe2O3 × (1 mol Fe2O3 / 159.7 g Fe2O3) = 0.626 moles Fe2O3.
Using the mole ratio from iron to iron(III) oxide (4:2), we can find the moles of Fe:
0.626 moles Fe2O3 × (4 moles Fe / 2 moles Fe2O3) = 1.252 moles Fe.
Finally, convert the moles of Fe to mass:
1.252 moles Fe × (55.85 g/mol Fe) = 69.9 g Fe.
Therefore, 69.9 g of iron is needed to form 100.0 g of rust.