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The rusting of iron is represented by the equation 4Fe + 3O2 --> 2Fe2O3. If you have a 3.5-mol sample of iron, how many grams of O2 are needed to rust the iron completely?

34.9 g
84.0 g
16.5 g
48.6 g

Sagot :

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Answer:

84.0 g

Balanced equation: 4Fe + 3O2 --> 2Fe2O3

  • 3.5-mol sample of iron

Find moles of oxygen:

  • 4Fe : 3O2
  • 4 : 3

  • (3.5/4)*3 = 2.625 moles

Find mass of O2

  • mass = moles * Mr
  • mass = 2.625 * 32
  • mass = 84 g
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