neutralisation reaction between iron and potassium manganate

To calculate the molecular mass i added up all the elements, (Fe(NH4)2 (SO4)2 6H2O), the molecular mass is 391.90

We can calculate the number of moles by using the following formula

Number of moles = concentration x volume

So 0.05 X 0.025(this has to be converted dm3) = 0.0125

APPARATUS:

• Volumetric flask
• 10 cm pipette
• 100 cm conical flask
• burette
• distilled water
• clamp stand
• standard solution
• Lab coats should be worn during the practical
• Eye protection should also be worn

METHOD

1. set up apparatus
2. fill burette with potassium manganate solution
3. get 250ml of the standard solution transfer it to a flask using a pipette
4. while swirling add potassium manganate to the solution
5. use a white tile to see a colour change (pink)
6. record the end point the first will be rough repeat method to get concordant results

Results Table

Average = 26.60cm3

5Fe2+(aq) + MnO4- (aq) + 8H+ →

 5Fe+3(aq)+Mn+2(aq) + 4H2O(l)

The reaction shows that 5 moles of iron II neutralises 1 mole of potassium mangnate

0.00125/5 = 0.00025

From this we can work out the no of moles of potassium mangnate

0.00025x1000/26.60 = 0.0093

To find the percentage of iron i used this equation

Moles of iron in 25cm3/moles in 25cm3XMr = 0.069/0.466=0.14263X100 = 14.8% i checked this value and compared my result to the original value of 14.3 and noticed i wasn’t far off