How I calculate the concentration of the limiting reactant solution? B=0.05MC=0.10MD30.05MID of Na3PO4 solution:ID of BaCl2 solution:Trial 1Trial 220.1 ML10.1 MLVolume of Na3PO4 solution:Volume of BaCl2 solution:20ML19.990,5419Mass of filter paper:Combined mass of paper and precipitate:1.135a1.3580.7399Mass of precipitate:Formula for limiting reactant: 2BachFormula for excess reactant:Calculated concentration of limiting reactant solution:005MReported concentration of excess reactant solution:0.05 MReported concentration of limiting reactant solution:(values will be provided during the lab period)105

Given: Precipitate: Ba3(PO4)2 molar mass of Ba3(PO4)2 = 601.93 g/mol

Trial 1 Trial 2 20.1 ML 20.1 ML Volume of Na3PO4 solution: Volume of BaCl2 solution: 20ML 1G.99 0,3|19 Mass of filter paper: Combined mass of paper and precipitate: 1.135 1.358 Mass of precipitate: Formula for limiting reactant: Formula for excess reactant: Calculated concentration of limiting reactant solution: 005M Reported concentration of excess reactant solution: 0.05 M Reported concentration of limiting reactant solution: (values will be provided during the lab period)

Trial 1 Trial 2 20.1 ML 20.1 ML Volume of Na3PO4 solution: Volume of BaCl2 solution: 20ML 1G.99 0,3|19 Mass of filter paper: Combined mass of paper and precipitate: 1.135 1.358 Mass of precipitate: Formula for limiting reactant: Formula for excess reactant: Calculated concentration of limiting reactant solution: 005M Reported concentration of excess reactant solution: 0.05 M Reported concentration of limiting reactant solution: (values will be provided during the lab period)

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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