The zinc in a 1.199 g sample of a foot powder was precipitated as ZnNH₄PO₄. Strong heating of the precipitate yielded 0.4089 g Zn₂P₂O₇. Calculate the mass percent of zinc in the sample of foot powder.\[ \boxed{\phantom{answer}} \, \% \]

Given : Mass of foot powder = 1.199 g Mass of Zn2P2O7 produced = 0.4089 g Mass of Zn2P2O7 produced = 0.4089 g Molar mass of Zn2P2O7 = 304.7 g/mol Atomic mass of Zinc = 65.38 g/mol Mass of 2 Zn = 2 × 65.38 = 130.76 g/mol Percent mass of Zinc in Zn2P2O7 = (130.76/ 304.7)× 100 = 42.9% Mass of zinc in Zn2P2O7 = ( mass percent of zinc/100) × total mass of Zn2P2O7 = (42.9 /100) × 0.4089 g = 0.1754g Mass of zinc will be same in foot powder as in Zn2P2O7 , hence mass of zinc = 0.1754 g Mass of foot powder given= 1.199g Mass percent of zinc in foot powder = ( mass of zinc/mass of foot powder) × 100 = (0.1754 g /1.199g) × 100 = 14.6% hence mass of zinc in foot powder= 14.6%

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Chemistry

The zinc in a 1.199 g sample of a foot powder was precipitated as ZNNH,PO4. Strong heating of the precipitate yielded 0.4089 g Zn,P207. Calculate the mass percent of zinc in the sample of Coot powder.

The zinc in a 1.199 g sample of a foot powder was precipitated as ZNNH,PO4. Strong heating of the precipitate yielded 0.4089 g Zn,P207. Calculate the mass percent of zinc in the sample of Coot powder.

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

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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