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?
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Chapter4: Chemical Reactions
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Problem 4.130QP: Arsenic acid, H3AsO4, is a poisonous acid that has been used in the treatment of wood to prevent...
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Transcribed Image Text:3. 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(aq) → 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 HC1?
(d) How many grams of Al(OH)3 reacted? The molar mass of Al(OH)3 is 78.00 g/mol.
Hint: You determined the number of moles of Al(OH)3 that reacted in (c) above.
(e) Given the mass of sample of suspension titrated (see problem statement above), what is the mass % of Al(OH)3 in that
suspension?
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