If 7.02 g of potassium dichromate (294.185 g/mol) is used to prepare a solution and then...

If 7.02 g of potassium dichromate (294.185 g/mol) is used to prepare a solution and then reacted with 18.15 g of an iron (II) compound. What mass of iron(II) is present in the sample?

Expert Solution

Given data:

The mass of potassium dichromate (K₂Cr₂O₇) used = 7.02 g.

The molar mass of K₂Cr₂O₇ = 294.185 g/mol

Therefore, the no. of moles of K₂Cr₂O₇ = 7.02/294.185, i.e. 0.02386

The mass of iron (II) compound used in the reaction = 18.15 g.

The molar mass of iron = 55.845 g/mol

Therefore, the no. of moles of Fe(II) = 18.15/55.845, i.e. 0.325

According to the balanced equation for the reaction between K₂Cr₂O₇ and Fe(II), there will be 1 (K₂Cr₂O₇) : 6 (Fe(II))-mole ratio.

Reason: Cr(VI) reduces to Cr(III). Here 2 Cr(VI) are present, so they need 6 electrons, whereas, Fe(II) oxidizes to Fe(III). Hence 6 moles of Fe(II) is needed for 1 mole of Cr(VI)

Therefore, the no. of moles of iron(II) present in the sample = 0.325 - (6*.02386), i.e. 0.18184

Hence, the mass of iron(II) present in the sample = 0.18184*55.845, i.e. 10.15 g.

queen_honey_blossom answered 2 years ago

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