Salt mixture: 2.64 g Mass of filter paper: 1.223 g Mass of filter paper and CaC2O4⋅H2O: 1.997 g Mass of air-dried CaC2O4⋅H2O: 0.774g 4. What is the mass percent of your limiting reactant? 5. What is the mass percent of your excess reactant?
Salt mixture: 2.64 g Mass of filter paper: 1.223 g Mass of filter paper and CaC2O4⋅H2O: 1.997 g Mass of air-dried CaC2O4⋅H2O: 0.774g 4. What is the mass percent of your limiting reactant? 5. What is the mass percent of your excess reactant?
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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Question
Salt mixture: 2.64 g
Mass of filter paper: 1.223 g
Mass of filter paper and CaC2O4⋅H2O: 1.997 g
Mass of air-dried CaC2O4⋅H2O: 0.774g
4. What is the mass percent of your limiting reactant?
5. What is the mass percent of your excess reactant?
Transcribed Image Text:According-
to the gruen informmeition we can sciy that -
Nassof le e Oq.Hz0 + Mass of eas O4: HzO+mass of filter
paper
= 2-64 gm
or nnass ofGG04.1+20 = 2.64-1.997
= 0.6439M
alsofromy -
massof filter þaper + nmass of CaçO4.lt20 = 1'997gmy
,. mass of caQ O4.H20 = l1997-1.22J
=017749un.
(utherefore nnass of linniting reagent kQ04. HeO in
the salt = o.6439M
(2) NO.of grcinns of Excess reactant Ca 204.1420
Inthe salt nmixture = 0,774
3) NO.of greinns of Exeess reactant left unreaeted
after the re ele tion is connpleted
= · 774 –1643
= 0•131gm
Note: Please repost 4 eind E þart again to be
answered.
Transcribed Image Text:@ aceording to the Question the
given
information
are-
Salt mixture =
Mass of filter þaper and CaG0q.l+20
massof dried caG04·It20 = 0.774 gm
2.64gm
1.997.gm
Zimiting reagent = k;G4. Hz O
nnass o f filter paper = 1· 223gm
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