What mass of iron(III) oxide is produced when 8 g of iron metal reacts completely with oxygen gas?
first, write the reaction equation.
Then figure out
(a) how many moles of Fe in 8g
(b) how many moles of Fe2O3 are produced for each mole of Fe used
(c) convert that back to grams
To determine the mass of iron(III) oxide produced when iron metal reacts completely with oxygen gas, we need to balance the chemical equation for the reaction and use stoichiometry.
First, let's write the balanced chemical equation for the reaction between iron metal (Fe) and oxygen gas (O₂) to form iron(III) oxide (Fe₂O₃):
4 Fe + 3 O₂ → 2 Fe₂O₃
The equation tells us that 4 moles of iron (Fe) react with 3 moles of oxygen gas (O₂) to form 2 moles of iron(III) oxide (Fe₂O₃).
Now, we need to calculate the molar mass of iron(III) oxide. The molar mass of Fe is 55.85 g/mol, and the molar mass of O is 16.00 g/mol.
For Fe₂O₃, we have:
2(Fe) + 3(O) = 2(55.85 g/mol) + 3(16.00 g/mol) = 159.70 g/mol
To determine the mass of iron(III) oxide produced, we can use the concept of stoichiometry.
1. Convert the mass of iron (Fe) to moles using its molar mass:
moles of Fe = 8 g / 55.85 g/mol ≈ 0.143 mol
2. Use the stoichiometric ratio from the balanced equation to relate the moles of iron to the moles of iron(III) oxide:
moles of Fe₂O₃ = (0.143 mol Fe) / (4 mol Fe₂O₃ / 3 mol Fe) ≈ 0.107 mol Fe₂O₃
3. Finally, convert the moles of iron(III) oxide to mass using its molar mass:
mass of Fe₂O₃ = 0.107 mol Fe₂O₃ × 159.70 g/mol ≈ 17.20 g
Therefore, when 8 g of iron metal reacts completely with oxygen gas, approximately 17.20 g of iron(III) oxide is produced.