Calculate the mass of Fe2O3 that contoins 1000 kg of iron
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Calculate the mass of Fe2O3 that contoins 1000 kg of iron

[From: ] [author: ] [Date: 12-10-17] [Hit: ]
1000kg x 1000g/1kg x 1 mol Fe2/ 111.69 g = 8953.8953.35 mol Fe2O3 x 159.69 g/ 1 mol Fe2O3x 1 kg/1000 g= 1.:) good luck,......
the book shows the answer in the back as 1.4 X 10^3 kg - i'm not even close.

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So....
1000kg x 1000g/1kg x 1 mol Fe2/ 111.69 g = 8953.35 mols of iron in Fe2O3

8953.35 mol Fe2O3 x 159.69 g/ 1 mol Fe2O3x 1 kg/1000 g= 1.43 x 10^3 kg of Fe2O3

So the steps to solving this problem are:
1) Find out how many mols 1000kg of Fe2 is
2) Since the number of mols of Fe2 is equal to the number of mols of Fe2O3 you have; use the formula weight of the Fe2O3 and the number of mols of Fe2O3 to solve for the weight in kg of Fe2O3
:) good luck, hope this helped
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