**Quantitative Analysis: Neutralization Reactions****1) Neutralizing Antacid Tablet:**An extra strength antacid tablet contains 750 mg of active ingredient, CaCO₃. If it takes 22.25 mL of HCl (aq) to neutralize the tablet, how strong is the acid?Chemical Reaction:\[ 2 \text{HCl (aq)} + \text{CaCO}_3\text{ (aq)} \rightarrow \text{CaCl}_2\text{ (aq)} + \text{HOH (l)} + \text{CO}_2\text{ (g)} \]**2) Neutralizing H₂SO₄ Spill:**A beaker containing 25.0 mL of 0.360 M H₂SO₄ spills on the counter. How much baking soda, NaHCO₃, will be needed to neutralize the acid?Chemical Reaction:\[ \text{H}_2\text{SO}_4\text{ (aq)} + 2 \text{NaHCO}_3\text{ (s)} \rightarrow \text{Na}_2\text{SO}_4\text{ (aq)} + 2 \text{H}_2\text{CO}_3\text{ (l)} \]**3) Neutralizing HCl Solution:**If a 35.75 mL sample of 0.175 M HCl (aq) solution is neutralized with 0.125 M Mg(OH)₂, how much Mg(OH)₂ will be needed?Chemical Reaction:\[ 2 \text{HCl (aq)} + \text{Mg(OH)}_2\text{ (aq)} \rightarrow \text{MgCl}_2\text{ (aq)} + 2 \text{HOH (l)} \]

Since you have posted multiple questions, we are entitled to answer the first only. 1) The balanced reaction taking place is given as, Given : Volume of HCl solution = 22.25 mL = 0.02225 L (since 1 L = 1000 mL) And mass of CaCO3 = 750 mg = 0.750 g (since 1 g = 1000 mg) Molar mass of CaCO3 = Atomic mass of Ca + Atomic mass of C + Atomic mass of O X 3 = 40 + 12 + 16 X 3 = 100 g/mol Since mass = moles X molar mass => 0.750 = moles of CaCO3 reacting X 100 => Moles of CaCO3 reacting = 7.50 X 10-3 molFrom the above balanced reaction we can see that 1 mole of CaCO3 is reacting with 2 moles of HCl. Hence moles of HCl present = moles of CaCO3 reacting X 2 = 7.50 X 10-3 X 2 = 0.0150 mol Since moles of HCl present = concentration of HCl X volume of HCl solution in L => 0.0150 = concentration of HCl solution X 0.02225 => Concentration of HCl solution = 0.674 M approx. Hence the strength of the HCl solution is 0.674 M.

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Chemistry

1) An extra strength antacid tablet contains 750 mg of active ingredient, CaCO3. If it takes 22.25 mL of HCI (aq) to neutralize the tablet, how strong is the acid? 2 HCI (aq) + CaCO; (ag) → CaCl2(aq) + HOH (1) + CO2 (g)

1) An extra strength antacid tablet contains 750 mg of active ingredient, CaCO3. If it takes 22.25 mL of HCI (aq) to neutralize the tablet, how strong is the acid? 2 HCI (aq) + CaCO; (ag) → CaCl2(aq) + HOH (1) + CO2 (g)

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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