Question

A 0.649 g sample containing only K2SO4 (174.27 g/mol) and (NH4)2SO4 (132.14 g/mol) was dissolved in water and treated with Ba(NO3)2 to precipitate all sulfate as BaSO4 (233.39 g/mol). If 0.977 g of precipitate was formed, what is the mass percent K2SO4 in the sample?

Name: a 0649 g sample containing only k2so4 17427 gmol and nh42so 13214 gmol was dissolved in water and treated with bano32 to precipitate all sulfate as baso4 23339 gmol if 0977 g of precipitate 27873
Uploaded: 2022-09-02T13:53:51-08:00
Duration: 5 min 9 s
Channel: Adi S
Description: a 0649 g sample containing only k2so4 17427 gmol and nh42so 13214 gmol was dissolved in water and treated with bano32 to precipitate all sulfate as baso4 23339 gmol if 0977 g of precipitate 27873

          A 0.649 g sample containing only K2SO4 (174.27 g/mol) and (NH4)2SO4 (132.14 g/mol) was dissolved in water and treated with Ba(NO3)2 to precipitate all sulfate as BaSO4 (233.39 g/mol). If 0.977 g of precipitate was formed, what is the mass percent K2SO4 in the sample?

Added by Patricia B.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Adi S

09/02/2022

Step 1

Given: - Total mass of the sample = 0.649 g - Molar mass of K2SO4 = 174.27 g/mol Let the mass of K2SO4 in the mixture be x g. Therefore, the mass of (NH4)2SO4 in the mixture = 0.649 - x g Moles of K2SO4 = x g / 174.27 g/mol = 0.00574x mol Show more…

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A 0.649 g sample containing only K2SO4 (174.27 g/mol) and (NH4)2SO4 (132.14 g/mol) was dissolved in water and treated with Ba(NO3)2 to precipitate all sulfate as BaSO4 (233.39 g/mol). If 0.977 g of precipitate was formed, what is the mass percent K2SO4 in the sample?