Please help with question 2 1. If you had to make a 0.750M solution of Na₂CO3 from 100mL of solution, determine themass of Sodium Carbonate necessary.100mL1000 mi/LX, 750 mol/L = .075 molNa₂CO3-1069/molmass of Na₂CO3 = 106g/molx.075md=2. If you had to make a 0.250M solution of CaCl₂ from 100mL of solution, determine themass of Calcium Chloride necessary.250m1000 meX100ML =·025ML of3. If you use 10.0 mL of the Sodium Carbonate solution and an excess of Calcium Chloride,how many grams of Calcium Carbonate should be theoretically be produced?10.0mL Mograms Na203 +CaCl2 +CaCO3+ 2NaCl.750M4. If you use 25.0 mL of the Calcium Chloride solution and an excess of Sodium Carbonate,how many grams of Calcium Carbonate should be theoretically be produced?5. Using the information in problems 3 and 4, what is the theoretical yield of 25.0mL of a0.250M solution of CaCl2 when mixed with 10.0 mL of a 0.750M Sodium Carbonatesolution?6. Suppose the student performs the experiment in the previous problem, what is thepercent yield if they generate 0.565g of CaCO3?

Molarity is a concentration term for a solution. The molarity of the given solution is defined as…

Answered: Na₂ LU 1069/m01 2. If you had to make a…

Na₂ LU 1069/m01 2. If you had to make a 0.250M solution of CaCl2 from 100mL of solution, determine the mass of Calcium Chloride necessary. 250m 1000 me X100ML = ·025ML of

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

Chapter4: Types Of Chemical Reactions And Solution Stoichiometry

Section: Chapter Questions

Problem 47E

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Concept explainers

Solutions

Solutions can be considered as homogeneous mixtures which contain two or more than two chemical substances. The chemical substance which is the greater part is defined as solvent. And the chemical substance that is dissolved in a solvent is referred to as solute. Solution chemistry plays an important role in chemistry since it supports many commercial applications of solutions.

Question

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Please help with question 2

1. If you had to make a 0.750M solution of Na₂CO3 from 100mL of solution, determine the
mass of Sodium Carbonate necessary.
100mL
1000 mi/LX, 750 mol/L = .075 mol
Na₂CO3-1069/mol
mass of Na₂CO3 = 106g/molx.075md=
2. If you had to make a 0.250M solution of CaCl₂ from 100mL of solution, determine the
mass of Calcium Chloride necessary.
250m
1000 me
X100ML =
·025ML of
3. If you use 10.0 mL of the Sodium Carbonate solution and an excess of Calcium Chloride,
how many grams of Calcium Carbonate should be theoretically be produced?
10.0mL Mograms Na203 +CaCl2 +CaCO3+ 2NaCl
.750M
4. If you use 25.0 mL of the Calcium Chloride solution and an excess of Sodium Carbonate,
how many grams of Calcium Carbonate should be theoretically be produced?
5. Using the information in problems 3 and 4, what is the theoretical yield of 25.0mL of a
0.250M solution of CaCl2 when mixed with 10.0 mL of a 0.750M Sodium Carbonate
solution?
6. Suppose the student performs the experiment in the previous problem, what is the
percent yield if they generate 0.565g of CaCO3?

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