Pre-lab: Iren Determinatiern 5. How many total sumber of moles of Fe3 are in the

Question

Pre-lab: Iren Determinatiern 5. How many total sumber of moles of Fe3 are in the iron tablet (One iroa tablet was dissolved in 250 ml, so multiply your previous answer by five) Data Mass of iroa tablet: 0.750 grams Concentration of the KMoO: 0.00400 Molas 37.0 35.0 6. How many moles of Feso. are in your iron tablet MnO in buret (mL) 360 30 2.0 7. What is the mass of Feso, are in your inon tablet? MaO, used (mL) Mno, used 8What is the percent of FeSo, in one iron tablet? Calculations 1. What is the average velume of MnO." used in units of liters 2. What is the average moles of Mao, used? What was the maln objeetive of this lab usingacouple of seatences . What is the belanced reaction whes reacting the iron with the permanganate in this lab? tthe equation given ln the introduction needs to be balancod) 4. ilow many moles of Fe ar·in your 50 s ..mpler

Explanation / Answer

1.

Volume = Final - initial reading

Trial 1: 37.0-3.0 = 34.0 mL;

Trial 2: 35.0-2.0 = 33.0 mL;

Trial 3: 42.0-7.0 = 35.0 mL;

Trial 4: 36.0-2.0 = 34.0 mL

Average volume = (34.0 + 33.0 +35.0 +34.0 )/ 4

= 34.0 mL

In Liters, volume =34.0 mL * (1 L/ 1000 mL)

= 0.034 L

2. Molarity of KMnO4 = 0.00400 M

Volume = 0.034 L

Average moles = molarity * volume

= 0.00400 M * 0.034 L

= 0.000136 moles

3.

The balanced equation for the reaction is:

5 Fe^2+ + MnO4^- + 8 H^+ ----> 5 Fe^3+ + Mn^2+ + 4 H2O

4.

1 mole of MnO4^- reacts with 5 moles of Fe2+

So moles of Fe^2+ = 5 * 0.000136 = 0.00068 moles

5.

Multiplying the previous answer by 5 gives moles of Fe^2+ in one tablet:

Moles of Fe^2+ = 0.00068 * 5

= 0.0034 moles

6.

One mole of FeSO4 contains 1 mole of Fe^2+. So moles of both are equal

So moles of FeSO4 = 0.0034 moles

7. Mass of FeSO4 = 0.0034 * (151.91 g/ 1 mole)

= 0.516 g

8.

Mass of iron tablet = 0.750 g

Mass % of FeSO4 = (mass of FeSO4 / mass of tablet) * 100

= (0.516 g / 0.750 g) * 100

= 68.8 %

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