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Calculate the mass of pure sample (NH4)2SO4 when it is dissolved in water and treated with excess NaOH using the balanced equations below: (NH4)2SO4 + 2NaOH ---> 2NH4OH + Na2SO4 2NH4OH ---> 2NH3 + H2O The liberated NH3 is passed into 50.0 mL of HCl. The excess acid (back titration) then requires 15.0 mL of 0.25 N KOH solution for titration. The HCl has a Normality of 0.2750 N. MM of (NH4)2SO4 = 132.14 g/mol

Calculate the mass of pure sample (NH4)2SO4 when it is dissolved in water and treated with excess NaOH using the balanced equations below: (NH4)2SO4 + 2NaOH ---> 2NH4OH + Na2SO4 2NH4OH ---> 2NH3 + H2O The liberated NH3 is passed into 50.0 mL of HCl. The excess acid (back titration) then requires 15.0 mL of 0.25 N KOH solution for titration. The HCl has a Normality of 0.2750 N. MM of (NH4)2SO4 = 132.14 g/mol

Principles of Modern Chemistry
Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter15: Acid–base Equilibria
Section: Chapter Questions
Problem 57P
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Calculate the mass of pure sample (NH4)2SO4 when it is dissolved in water and treated with excess NaOH using the balanced equations below:

(NH4)2SO4 + 2NaOH ---> 2NH4OH + Na2SO4

2NH4OH ---> 2NH3 + H2O

The liberated NH3 is passed into 50.0 mL of HCl. The excess acid (back titration) then requires 15.0 mL of 0.25 N KOH solution for titration. The HCl has a Normality of 0.2750 N.

MM of (NH4)2SO4 = 132.14 g/mol
MM of Na2SO4 = 142.04 g/mol
MM of HCl = 36.46 g/mol
MM of NaOH = 40.00 g/mol
MM of NH4OH = 35.04 g/mol
MM of NH3 = 17.03 g/mol
MM of KOH = 56.11 g/mol

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