a solution of 0.010M K2Cr2O7 with a volume of 60.00ml reacts in a titration reaction with 20.00ml of FeSO4 in sulphuric acid. Calculate the normality of FeSO4.
2 K2Cr2O7 + 3 FeSO4 + 14 H2SO4 = 2 K2SO4 + 2 Cr2(SO4)3 + 3 Fe(SO4)3 + 14 H2O
for every 2 mols of K2Cr2O7 you need 3 mols of FeSO4
You have .01*60*10^-3 mols of K2Cr2O7
so you need
.01(3/2)*60*10^-3 mols of Fe(SO4)3
in 20*10^-3 liters
.01(1.5)60/20 = .045 mols/liter
6 FeSO4 + K2Cr2O7 + 7 H2SO4 → K2SO4 + 3 Fe2[SO4]3 + Cr2[SO4]3 + 7 H2O
shouldn't this be the balanced chemical equation?
making it for every mole of K2Cr2O7 you need 6 moles of FeSO4??? and calculations can proceed from this information
To calculate the normality of FeSO4, we need to determine the number of moles of FeSO4 involved in the reaction.
First, let's write the balanced chemical equation for the titration reaction between potassium dichromate (K2Cr2O7) and iron(II) sulfate (FeSO4):
K2Cr2O7 + 6FeSO4 + 7H2SO4 → 3Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + 7H2O
From the balanced equation, we can see that the mole ratio between K2Cr2O7 and FeSO4 is 1:6. In other words, one mole of K2Cr2O7 reacts with six moles of FeSO4.
Given that the concentration of K2Cr2O7 is 0.010 M and the volume used is 60.00 ml (or 0.060 L), we can calculate the number of moles of K2Cr2O7:
moles of K2Cr2O7 = concentration × volume
= 0.010 M × 0.060 L
= 0.0006 moles
Since the mole ratio between K2Cr2O7 and FeSO4 is 1:6, the number of moles of FeSO4 involved in the reaction is:
moles of FeSO4 = 6 × moles of K2Cr2O7
= 6 × 0.0006 moles
= 0.0036 moles
Next, we need to determine the volume of FeSO4 solution used in the titration, which is given as 20.00 ml (or 0.020 L).
Using the volume and the number of moles of FeSO4, we can calculate its normality:
normality of FeSO4 = moles of FeSO4 / volume in liters
= 0.0036 moles / 0.020 L
= 0.18 N
Therefore, the normality of FeSO4 in this titration reaction is 0.18 N.