2.) The first step in the reaction of Alka-seltzer with Stomach acid consists of one reacting with one more of (HCI) one mole of (HCI) OF CH₂ CO3), and one mole of mole of (NaHCO3) to produce one mole (NaCl) O using this Stoichiometry, determine the number of moles Of Carbonic acid that can be produced from 5 mol NaHCO3 and 8 mol HCl 2 Which of the two reactants limit the moles OF H₂ CO3 that can be made ? How much excess reactant remains after the reaction?

2.) The first step in the reaction of Alka-seltzer with Stomach acid consists of one reacting with one more of (HCI) one mole of (HCI) OF CH₂ CO3), and one mole of mole of (NaHCO3) to produce one mole (NaCl) O using this Stoichiometry, determine the number of moles Of Carbonic acid that can be produced from 5 mol NaHCO3 and 8 mol HCl 2 Which of the two reactants limit the moles OF H₂ CO3 that can be made ? How much excess reactant remains after the reaction?

Introductory Chemistry: A Foundation

9th Edition

ISBN:9781337399425

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Donald J. DeCoste

Chapter9: Chemical Quantities

Section: Chapter Questions

Problem 100CP: he production capacity for acrylonitrile (C3H3N)in the United States is over 2 billion pounds per...

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Consider Palmitic acid C16H32O2, a common fatty acid used in the manufacture of soap. A solution of palmitic acid is prepared by mixing 112 g palmitic acid with 725 mL of benzene C6H6. The density of the resulting solution is 0.902 g/mL. Palmitic acid (molar mass = 256 g/mol and density = 0.852 g/mL); Benzene (molar mass = 78 g/mol and density = 0.879 g/mL) A. What is the %m/m of the solution? B. What is the molarity of the solution? C. What is the mole fraction of the solvent?
Consider Palmitic acid C16H32O2, a common fatty acid used in the manufacture of soap. A solution of palmitic acid is prepared by mixing 112 g palmitic acid with 725 mL of benzene C6H6. The density of the resulting solution is 0.902 g/mL. Palmitic acid (molar mass = 256 g/mol and density = 0.852 g/mL); Benzene (molar mass = 78 g/mol and density = 0.879 g/mL) 1. What is the % m/m of the solution? 2. What is the mole fraction of the solvent? 3. What is the molarity of the solution? Based from the calculated molarity of the solution in #3, you are to prepare 250 mL of the said molar concentration. If you have an available 2M stock solution of Palmitic acid, How many mL of the 2M palmitic acid stock solution will you need? and; 4. How many mL of solvent do you need to add to the amount gotten from the 2M Palmitic acid stock solution in order to prepare 250 mL total volume of your desired solution?