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3.47 g of the hydrated "double salt", ammonium iron(II) sulfate hexahydrate, FeSO4(NH4)2SO4•6H2O was

3.47 g of the hydrated "double salt", ammonium iron(II) sulfate hexahydrate, FeSO4(NH4)2SO4•6H2O was

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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3.47 g of the hydrated "double salt", ammonium iron(II) sulfate hexahydrate, FeSO4(NH4)2SO4-6H2O was dissolved in 200. mL of water. 20.0 mL of the solution had some acid added to it and then it reacted completely with 12.6 mL of KMNO4 solution. Calculate the concentration of the KMNO4 solution given the full REDOX equation below. (4) 5FE2* + MnO4 + 8H* → 5FE3* + Mn2+ + 4H2O
Transcribed Image Text:3.47 g of the hydrated "double salt", ammonium iron(II) sulfate hexahydrate, FeSO4(NH4)2SO4-6H2O was dissolved in 200. mL of water. 20.0 mL of the solution had some acid added to it and then it reacted completely with 12.6 mL of KMNO4 solution. Calculate the concentration of the KMNO4 solution given the full REDOX equation below. (4) 5FE2* + MnO4 + 8H* → 5FE3* + Mn2+ + 4H2O
Expert Solution
Step 1 Redox reactions

1. Molarity of salt :

M = mole of salt / Volume of solution (in L)

M = n/V (in L)

where n = mass of salt /molar mass of salt

mass of salt = 3.47 g, Molar mass of salt = 392.13 g/mol,

Volume = 200 ml x 1L/1000mL =0.2 L

there fore 

Msalt  = (3.47 g / 392.13 g/mol)   x  1/ 0.2 L

          = 0.044 M

In redox reaction Fe+2 is converted into Fe+3  Change in oxidation number of Fe is 1 (1 unit increased)

Hence valence factor is 1

there fore 

Normality of salt (Nsalt) = Msalt  x Valence factor

                                       = 0.044  x 1

                                       = 0.044 N

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