Data Mass of iron tablet: 0.750 grams Concentration of the KMno4: 0.00400 Molar

Question

Data Mass of iron tablet: 0.750 grams Concentration of the KMno4: 0.00400 Molar Trial #1 Trial #2 Trial #3 Trial #4 Final reading of 35.0 36.0 37.0 42.0 Mno41 in buret (mL) Initial reading 3.0 2.0 2.0 of Mno4-1 in buret Volume of Mno41 used Volume of Mno41 used Calculations 1. What is the average volume of Mno4 used in units of liters? 2. What is the average moles of Mno4-1 used 3. What is the balanced reaction when reacting the iron with the permanganate in this lab? (the equation given in the introduction needs to be balanced 4. How many moles of Fe are in your 50 mL sample)? +2

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 %

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.