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
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
Chapter14: Acids And Bases
Section: Chapter Questions
Problem 194IP: Isocyanic acid (HNCO) can be prepared by heating sodium cyanate in the presence of solid oxalic acid...
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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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