This is 1 question **Titration of Aluminum Hydroxide Suspension**A 1.775 g aqueous suspension containing Al(OH)₃ (and water) was titrated with a solution containing 0.2055 moles of HCl per liter of solution (0.1055 M). It took 32.87 mL of the HCl solution to just react with all of the Al(OH)₃.**(a) Determine the Volume of HCl Solution Needed**- What is the volume of the HCl solution needed to react with all of the Al(OH)₃, in L? - Hint: Convert mL to L using 1 mL = 10⁻³ L.**(b) Calculate the Moles of HCl Needed**- How many moles of HCl were needed to react with all the Al(OH)₃? - Hint: Use the volume in liters and the concentration of HCl.**(c) Balanced Reaction Equation**- The balanced equation for the chemical reaction is: \[ \text{Al(OH)}_3(s) + 3 \text{HCl(aq)} \rightarrow \text{AlCl}_3(aq) + 3 \text{H}_2\text{O}(l) \]- From this, 1 mole of Al(OH)₃ reacts with 3 moles of HCl. Using your result from (b), determine how many moles of Al(OH)₃ reacted.**(d) Calculate the Mass of Al(OH)₃ Reacted**- How many grams of Al(OH)₃ reacted? Given the molar mass of Al(OH)₃ is 78.00 g/mol.**(e) Determine Mass Percent of Al(OH)₃ in Suspension**- What is the mass percentage of Al(OH)₃ in the suspension, based on the original mass of 1.775 g?**Note:** Each step builds on the information of the previous, using stoichiometry and conversion factors to find the necessary quantities.

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A 1.775 g-aqueous suspension containing Al(OH)3(s) (and water!) was titrated with a solution containing 0.2055 moles of HCl(ag) per L of solution (i.e., 0.1055 M). It took 32.87 mL of the HCl(aq) solution to just react with all of the Al(OH)3. (a) What is the volume of the HCl solution needed to react with all of the Al(OH)3, in L? Hint: This volume was given in the problem already, but in units of mL rather than L. Just convert using 1 mL = 10³L (b) How many moles of HCl were needed to react with all the Al(OH)3? Hint: The solution contained 0.2055 moles of HCl in each L, and you just determined in (a) how many liters of this solution were needed. (c) The balanced equation for the chemical reaction between Al(OH)3 and HCl is: Al(OH)3(s) + 3 HCl(ag) → AlCl3(aq) + 3 H2O() This means that only 1 mole of Al(OH)3 reacts for every 3 moles of HCl that react. Given that the number of moles of HCl that reacts is your answer to (b) above, how many moles of Al(OH)3 must have reacted with the HCl?

A 1.775 g-aqueous suspension containing Al(OH)3(s) (and water!) was titrated with a solution containing 0.2055 moles of HCl(ag) per L of solution (i.e., 0.1055 M). It took 32.87 mL of the HCl(aq) solution to just react with all of the Al(OH)3. (a) What is the volume of the HCl solution needed to react with all of the Al(OH)3, in L? Hint: This volume was given in the problem already, but in units of mL rather than L. Just convert using 1 mL = 10³L (b) How many moles of HCl were needed to react with all the Al(OH)3? Hint: The solution contained 0.2055 moles of HCl in each L, and you just determined in (a) how many liters of this solution were needed. (c) The balanced equation for the chemical reaction between Al(OH)3 and HCl is: Al(OH)3(s) + 3 HCl(ag) → AlCl3(aq) + 3 H2O() This means that only 1 mole of Al(OH)3 reacts for every 3 moles of HCl that react. Given that the number of moles of HCl that reacts is your answer to (b) above, how many moles of Al(OH)3 must have reacted with the HCl?

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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