1. A solution of 0.010 M K2Cr2O7 with a volume of 60.00 mL reacts in a titration reaction
with 20.00 mL of FeSO4 in sulphuric acid. Calculate the normality of FeSO4. 6Fe
2+ + Cr2O7
2- + 14 H
6Fe^2+ + Cr2O7^2- + 14H^+ ==> 2Cr^3+ + 6Fe^3+ + 7H2O
millimoles K2Cr2O7 = mL x M = 60.00 x 0.01 = 0.6000
Convert to mmoles Fe that is 0.6000 x (6 mols FeSO4/1 mol K2Cr2O7) = 3.6
Then M FeSO4 = mmols/mL = ?
You should add the zeros and watch the number of significant figures.
To calculate the normality of FeSO4, we need to first find the number of moles of FeSO4 that reacted in the titration.
Given:
Concentration of K2Cr2O7 solution (c1) = 0.010 M
Volume of K2Cr2O7 solution (V1) = 60.00 mL = 0.0600 L
Volume of FeSO4 solution (V2) = 20.00 mL = 0.0200 L
The balanced equation for the reaction is:
6Fe2+ + Cr2O72- + 14H+ -> 6Fe3+ + 2Cr3+ + 7H2O
From the balanced equation, we can see that the stoichiometric ratio between Fe2+ and Cr2O72- is 6:1. Therefore, the moles of Cr2O72- used in the reaction is given by:
moles of Cr2O72- = concentration x volume = c1 x V1 = 0.010 M x 0.0600 L = 0.0006 mol
Since the stoichiometric ratio between Fe2+ and Cr2O72- is 6:1, the moles of Fe2+ used in the reaction is 6 times the moles of Cr2O72-:
moles of Fe2+ = 6 x moles of Cr2O72- = 6 x 0.0006 mol = 0.0036 mol
Now, we can calculate the normality of FeSO4 using the formula:
Normality = (moles of solute) / (volume of solution in liters)
Normality of FeSO4 = moles of Fe2+ / V2 = 0.0036 mol / 0.0200 L = 0.18 N
Therefore, the normality of FeSO4 solution is 0.18 N.