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What mass of rust (Fe2O3) contains 25.0 g of Fe?
DrBob222
answered3 years ago
Anonymous
answered2 years ago
56
Explain Bot
answered7 months ago
To determine the mass of rust (Fe2O3) containing 25.0 g of Fe, we need to use the molar ratio between Fe and Fe2O3.
The molar ratio is determined by the stoichiometry of the balanced chemical equation for the reaction between Fe and O2, which forms Fe2O3.
The balanced equation is:
4 Fe + 3 O2 → 2 Fe2O3
From the equation, we can see that for every 4 moles of Fe, we get 2 moles of Fe2O3.
Now, we need to calculate the number of moles of Fe using the given mass. We can do this by dividing the mass (in grams) by the molar mass of Fe. The molar mass of Fe is approximately 55.85 g/mol.
Number of moles of Fe = mass of Fe / molar mass of Fe
Number of moles of Fe = 25.0 g / 55.85 g/mol
Number of moles of Fe ≈ 0.447 mol
Using the molar ratio, we can now determine the number of moles of Fe2O3.
Number of moles of Fe2O3 = (Number of moles of Fe / 4) * 2
Number of moles of Fe2O3 = (0.447 mol / 4) * 2
Number of moles of Fe2O3 ≈ 0.2235 mol
Finally, we can calculate the mass of Fe2O3 using the number of moles and the molar mass of Fe2O3. The molar mass of Fe2O3 is approximately 159.69 g/mol.
Mass of Fe2O3 = Number of moles of Fe2O3 * molar mass of Fe2O3
Mass of Fe2O3 = 0.2235 mol * 159.69 g/mol
Mass of Fe2O3 ≈ 35.7 g
Therefore, the mass of rust (Fe2O3) that contains 25.0 g of Fe is approximately 35.7 grams.