Suppose that 25 mL of an iron (II) nitrate solution is added to a beaker contain

Question

Suppose that 25 mL of an iron (II) nitrate solution is added to a beaker containing Aluminum metal. Assume the reaction went to completion with no excess reagents. The solid iron produced was removed by filtration.

Write a balanced redox equation for the reaction.

For which solution, the initial iron (II) nitrate or the solution after the solid iron was filtered out, would be a better electrolyte? Explain your answer. Calculate the molarity of the solution if 1.00 g of Iron was produced.

Please explain step by step, thank you!!

Explanation / Answer

The balanced equation is-

3Fe(NO3)2 + 2Al = 3Fe + 2Al(NO3)3

Ionization of electrolytes

Al(NO3)3(aq) -------> Al3+(aq) + 3NO3-(aq)
x M ........................... x M ............ 3x M
Total ion concentration : 4x M

Total ion concentration in the solution is more after the precipitation of solid iron, so it will be a better electrolyte; more number of ions, better the electric conduction.

Fe(NO3)2(aq) -------> Fe2+(aq) + 2NO3-(aq)
x M ........................... x M ............ 2x M
Total ion concentration : 3x M

Molar mass of Fe = 55.845 gm/mole

Molar mass of Fe(NO3)2 is 179.8548 g/mol

3*55.845 g of Fe is produced from 3*179.8548 g of Fe(NO3)2.

1 g is produced from =3*179.8548/3*55.845/g of Fe(NO3)2 = 3.22 g of Fe(NO3)2

Therefore, molarity of the 25 ml solution = 3.22*1000/25/179.8548 = 0.716 M

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