Chemistry Problem?

[匿名]

2016-04-26 7:11 am

更新1:

The titration of 25.0 mL of an iron(II) solution required 18.0 mL of a 0.100 M solution of dichromate to reach the equivalence point. What is the molarity of the iron(II) solution?

更新2:

How do I solve this question? I tried M1V1=M2V2 and plugged in (.100mol/L)(.0250)=M2(.018L) and got .13889M as the answer, but it was wrong.

回答 (1)

不用客氣

2016-04-26 7:31 am

Half equation of reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Half equation of oxidation: Fe²⁺ → Fe³⁺ + e⁻

Complete equation : Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺
1 mole of Cr₂O₇²⁻ reacts with 6 moles of 6Fe²⁺.

Number of milli-moles of Cr₂O₇²⁻ reacts = (0.100 mmol/mL) × (18.0 mL) = 1.80 mmol
Number of milli-moles of Fe²⁺ reacts = (1.80 mmol) × 6 = 10.8 mmol
Molarity of Fe²⁺ = (10.8 mmol) / (25.0 mL) = 0.432 mol/L

👍 0 👎 0