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The sodium bicarbonate NaHCO3, sample is only 72.56% pure after it has been analyzed by the student chemist. The burette was filled with 0.22 M HCI up to 0.52 ml one- 2 drops of the indicator were added to the dissolved sample and were then titrated with 0.24 M HCI until the endpoint has been observed. The final reading was 13.18 mL 1. Solve for the molecular weight of the sample. Mass H=1, Na=23, O=16, C=12 Na H = 1X1 C = 12x1 = 23x1 =23 =1 =12 O 16x3 =4 2. Solve for the net volume of 0.22 M HCI in Liters ] = 84 g/mol NaHCO3 3. What is the mole ratio of sodium bicarbonate to HCI

The sodium bicarbonate NaHCO3, sample is only 72.56% pure after it has been analyzed by the student chemist. The burette was filled with 0.22 M HCI up to 0.52 ml one- 2 drops of the indicator were added to the dissolved sample and were then titrated with 0.24 M HCI until the endpoint has been observed. The final reading was 13.18 mL 1. Solve for the molecular weight of the sample. Mass H=1, Na=23, O=16, C=12 Na H = 1X1 C = 12x1 = 23x1 =23 =1 =12 O 16x3 =4 2. Solve for the net volume of 0.22 M HCI in Liters ] = 84 g/mol NaHCO3 3. What is the mole ratio of sodium bicarbonate to HCI

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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The sodium bicarbonate NaHCO3, sample is only 72.56% pure after it has been analyzed by the student chemist. The burette was filled with 0.22 M HCl up to 0.52 ml one- 2 drops of the indicator were added to the dissolved sample and were then titrated with 0.24 M HCI until the endpoint has been observed. The final reading was 13.18 mL 1. Solve for the molecular weight of the sample. Mass H=1, Na=23, O=16, C-12 Na H = 1X1 C = 12x1 =1 =12 O 10 16x3 =4 2. Solve for the net volume of 0.22 M HCI in Liters = 23x1 = 23 ?]. = 84 g/mol NaHCO3 3. What is the mole ratio of sodium bicarbonate to HCI 1:1 NaHCO3 + HCl NaCl + H₂ C03 4. Solve for the number of grams of sodium carbonate needed to get 83.68% 5. Solve for the number of moles sodium bicarbonate 6. Solve for the number of moles of HCI present in the mixture
Transcribed Image Text:The sodium bicarbonate NaHCO3, sample is only 72.56% pure after it has been analyzed by the student chemist. The burette was filled with 0.22 M HCl up to 0.52 ml one- 2 drops of the indicator were added to the dissolved sample and were then titrated with 0.24 M HCI until the endpoint has been observed. The final reading was 13.18 mL 1. Solve for the molecular weight of the sample. Mass H=1, Na=23, O=16, C-12 Na H = 1X1 C = 12x1 =1 =12 O 10 16x3 =4 2. Solve for the net volume of 0.22 M HCI in Liters = 23x1 = 23 ?]. = 84 g/mol NaHCO3 3. What is the mole ratio of sodium bicarbonate to HCI 1:1 NaHCO3 + HCl NaCl + H₂ C03 4. Solve for the number of grams of sodium carbonate needed to get 83.68% 5. Solve for the number of moles sodium bicarbonate 6. Solve for the number of moles of HCI present in the mixture
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