Answer:
To calculate the mass of Fe₂O₃ that contains 3.4×10³ kg of iron, we can use stoichiometry to find the molar mass of Fe₂O₃ and then calculate the mass of Fe₂O₃.
First, we need to find the molar mass of Fe₂O₃:
Molar mass of Fe = 55.85 g/mol
Molar mass of O = 16.00 g/mol
So, molar mass of Fe₂O₃ = 2(55.85) + 3(16.00) = 159.70 g/mol
Now we can calculate the mass of Fe₂O₃ that contains 3.4×10³ kg of iron:
Number of moles of Fe = mass of iron / molar mass of Fe
Number of moles of Fe = 3.4×10³ kg / 55.85 g/mol = 60809.037 moles
Since there are 2 moles of Fe for every 1 mole of Fe₂O₃, we can divide the number of moles of Fe by 2 to find the number of moles of Fe₂O₃:
Number of moles of Fe₂O₃ = 60809.037 moles / 2 = 30404.5185 moles
Finally, we can calculate the mass of Fe₂O₃:
Mass of Fe₂O₃ = number of moles of Fe₂O₃ × molar mass of Fe₂O₃
Mass of Fe₂O₃ = 30404.5185 moles × 159.70 g/mol = 4857981.1685 g
Converting this to kilograms:
Mass of Fe₂O₃ = 4857981.1685 g / 1000 = 4857.9811685 kg
So, the mass of Fe₂O₃ that contains 3.4×10³ kg of iron is approximately 4857.981 kg.
Explanation:
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Answer:
Explanation:
Molar mass of Fe2O3 =
2*55.845 * 3*15.999
= 159.687
So the answer, by proportion, is:
(2*55.845/ 159.687) * 3400
= 2378.065 kg
- that is about 2.378 * 10^3 kg of iron