LAST MINUTE QUESTIONS <3

[crayolé]

How would you do this one?
1.380g of an iron sa
lt was dissolved in 25.mL of water. The solution was titrated against .1 M Na2Cr2O7 and the titre was 12.78 mL. Using the half reaction Cr207 +14H +6e----> 2Cr 3+ + 7H2O
What was the formula of the iron salt?
Iron (II) Chloride
Iron (II) Sulfate
Iron (III) Chloride
Iron (II) Nitrate

[Potter]

Do the half eq of Fe2+ --> Fe3+ +e
multiply it by 6, to equate electrons

mol ratio it find, the mol of Fe2+ then 1.380/mol

should end up with iron nitrate.. Off the top of my head.

[scocliffe09]

How would you do this one?
1.380g of an iron sa
lt was dissolved in 25.mL of water. The solution was titrated against .1 M Na2Cr2O7 and the titre was 12.78 mL. Using the half reaction Cr207 +14H +6e----> 2Cr 3+ + 7H2O
What was the formula of the iron salt?
Iron (II) Chloride
Iron (II) Sulfate
Iron (III) Chloride
Iron (II) Nitrate

As Fe2+-->Fe3+ +e-
you can work out the ratio based on electrons.
i.e. 6Fe2+-->6Fe3+ +6e-
the rest is just mol calculations
n(Na2Cr2O7)=cV=0.1x0.01278=0.001278mol
n(Fe2+)=6x0.001278=0.007668mol
then n=m/M --> M=m/n
M=1.38/0.007668=180 --> iron nitrate