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4. If you use 25.0 mL of the Calcium Chloride solution from #2 and an excess of Sodium Carbonate, how many grams of Calcium Carbonate should be theoretically be produced? 5. Using your answers from 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?

4. If you use 25.0 mL of the Calcium Chloride solution from #2 and an excess of Sodium Carbonate, how many grams of Calcium Carbonate should be theoretically be produced? 5. Using your answers from 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?

Chemistry
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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Please help with questions 4, 5, and 6
Safety: While these are household items that you could easily purchase at the store, please understand that this is still for chemistry purposes. Do not INGEST anything or leave readily available for pets and young children to get into (especially your chalk, um-l mean your final precipitate). All items can be thrown away at the end. Ice melt solutions can be thrown away outside away from pet areas. Even though Calcium Carbonate is relatively harmless-please note that this should not be ingested. Prelab: 1. If you had to make a 0.750M solution of Na₂CO3 from 100mL of solution, determine the mass of Sodium Carbonate necessary. 1000 mL of 0.750m Solution contains 0.75 mol of Na₂CO3 100 mL of 0:750m Solution = 0.750x100 = 0.075 mol Na₂CO3 molar mass of Na₂CO3 = 105.988 g/m0000 0.075 mol NaCO3 =0.075x105.988 27.949 grams of Na₂CO3 2. If you had to make a 0.250M solution of CaCl₂ from 100mL of solution, determine the mass of Calcium Chloride necessary. 1000 mL 0.250m Solution contains 0.25 mol CaCl2 100 m² of 0.250m solution = 0.25×100 =0.025 mol of CaCl₂ 1000 mass of Sodium Carbonate= molar mass of CaCl₂=110.98g/mol 0.025 mol of CaCl2 = 0.025 x 110.98 = 2.77 grams of CaC1z mass of Calcium Chloride=; 7.949 grams e=2.77 grams 3. If you use 10.0 mL of the Sodium Carbonate solution from #1 and an excess of Calcium Chloride, how many grams of Calcium Carbonate should be theoretically be produced? Na₂CO3(aq) +CaCl₂ (aq) →→CaCo3 + 2NaCl(aq) 10 mL of 0.75 m Na₂CO3 = 0.75×10=0.00 75 mol Na₂CO3 0.75 grams of CaCo3 will be theoretically Produced Imol Na₂CO3 reacts w/ Callz producing I mol caco 3 molar mass of CaCo3 = 100.086 g/mol 0.0075 mol of Calo 320.00 75% 100.086 = 0.75 grams of CaCo3 4. If you use 25.0 mL of the Calcium Chloride solution from #2 and an excess of Sodium Carbonate, how many grams of Calcium Carbonate should be theoretically be produced? 5. Using your answers from 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?
Transcribed Image Text:Safety: While these are household items that you could easily purchase at the store, please understand that this is still for chemistry purposes. Do not INGEST anything or leave readily available for pets and young children to get into (especially your chalk, um-l mean your final precipitate). All items can be thrown away at the end. Ice melt solutions can be thrown away outside away from pet areas. Even though Calcium Carbonate is relatively harmless-please note that this should not be ingested. Prelab: 1. If you had to make a 0.750M solution of Na₂CO3 from 100mL of solution, determine the mass of Sodium Carbonate necessary. 1000 mL of 0.750m Solution contains 0.75 mol of Na₂CO3 100 mL of 0:750m Solution = 0.750x100 = 0.075 mol Na₂CO3 molar mass of Na₂CO3 = 105.988 g/m0000 0.075 mol NaCO3 =0.075x105.988 27.949 grams of Na₂CO3 2. If you had to make a 0.250M solution of CaCl₂ from 100mL of solution, determine the mass of Calcium Chloride necessary. 1000 mL 0.250m Solution contains 0.25 mol CaCl2 100 m² of 0.250m solution = 0.25×100 =0.025 mol of CaCl₂ 1000 mass of Sodium Carbonate= molar mass of CaCl₂=110.98g/mol 0.025 mol of CaCl2 = 0.025 x 110.98 = 2.77 grams of CaC1z mass of Calcium Chloride=; 7.949 grams e=2.77 grams 3. If you use 10.0 mL of the Sodium Carbonate solution from #1 and an excess of Calcium Chloride, how many grams of Calcium Carbonate should be theoretically be produced? Na₂CO3(aq) +CaCl₂ (aq) →→CaCo3 + 2NaCl(aq) 10 mL of 0.75 m Na₂CO3 = 0.75×10=0.00 75 mol Na₂CO3 0.75 grams of CaCo3 will be theoretically Produced Imol Na₂CO3 reacts w/ Callz producing I mol caco 3 molar mass of CaCo3 = 100.086 g/mol 0.0075 mol of Calo 320.00 75% 100.086 = 0.75 grams of CaCo3 4. If you use 25.0 mL of the Calcium Chloride solution from #2 and an excess of Sodium Carbonate, how many grams of Calcium Carbonate should be theoretically be produced? 5. Using your answers from 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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