Chemistry

Calculate the percentage of iron in Ferric oxide Fe₂O₃.

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Relative molecular mass of Ferric oxide (Fe₂O₃)
  = 56 x 2 + 16 x 3
  = 112 + 48
  = 160 amu

Since 160 g of Ferric oxide (Fe₂O₃ ) contains 112 g of iron

∴ 100 g of contains Ferric oxide (Fe₂O₃ ) contains $\dfrac{112}{160}$ x 100 = 70%

∴ Percentage of Iron in Ferric oxide (Fe₂O₃ ) is 70%

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