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Sagot :
Explanation:
Part a)
To find the percent by mass of Iron (III) oxide we have to suppose that we have 1 mol of the compound.
Formula of iron (III) oxide = Fe₂O₃
Molar mass of Fe = 55.85 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of Fe₂O₃ = 2 * 55.85 g/mol + 3 * 16.00 g/mol
Molar mass of Fe₂O₃ = 159.7 g/mol
If 1 mol of Fe₂O₃ molecules, we will have 2 moles of Fe atoms and 3 moles of O atoms. Let's find the mass of each element.
Mass of Fe₂O₃ sample = 1 mol of Fe₂O₃ * 159.7 g/mol = 159.7 g
Mass of Fe₂O₃ sample = 159.7 g
Mass of Fe = 1 mol of Fe₂O₃ * 2 moles of Fe/(1 mol of Fe₂O₃) * 55.85 g of Fe/(1 mol of Fe)
Mass of Fe = 111.7 g
Mass of O = 1 mol of Fe₂O₃ * 3 moles of O/(1 mol of Fe₂O₃) * 16.00 g of O/(1 mol of O)
Mass of O = 48.0 g
So in a 159.7 g sample of Fe₂O₃ we have 111.7 g of Fe and 48.0 g of O. Let's find their percent by mass.
% by mass of Fe = 111.7 g/(159.7 g) * 100
% by mass of Fe = 69.94 %
% by mass of O = 48.0 g/(159.7 g) * 100
% by mass of O = 30.06 %
Part b)
Now we have to find the mass of iron present in a 341.2 g sample of
Answer:
a) % by mass of Fe = 69.94 % % by mass of O = 30.06 %
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