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100.0 mL of HCl required 5.831 g Mg(OH)2 to titrate. The molar mass of Mg(OH)2 is 58.31 g. What is the concentration of the HCI ? The neutralization equation is 2 HCI + Mg(OH)₂ → MgCl₂ + 2 H₂O There is enough information to calculate the moles of Mg(OH)2 mol Mg(OH)2 (5.831 g Mg(OH)2 (- g Mg(OH)₂ Then use stoichiometry to convert moles of Mg(OH)2 into moles of HCI mol HCI concentration = mol Mg(OH)2 (-- a. acid i. 0.2000 p. 0.01000 d. NaOH j. 1.000 k. 0.005000 q. 0.001000 mol Mg(OH)2 We now can calculate the concentration of HCI since we have both the moles of HCI and the volume of HCI mol HCI b. base c. HCI )= LHCI r. 58.31 ) = e. NaCl I. 100.0 f. H₂O m. 0.1000 -) = s. 5.831 g. 25.00 mol Mg(OH)2 mol/L n. 0.05000 h. 0.02500 mol HCI o. 2.000

100.0 mL of HCl required 5.831 g Mg(OH)2 to titrate. The molar mass of Mg(OH)2 is 58.31 g. What is the concentration of the HCI ? The neutralization equation is 2 HCI + Mg(OH)₂ → MgCl₂ + 2 H₂O There is enough information to calculate the moles of Mg(OH)2 mol Mg(OH)2 (5.831 g Mg(OH)2 (- g Mg(OH)₂ Then use stoichiometry to convert moles of Mg(OH)2 into moles of HCI mol HCI concentration = mol Mg(OH)2 (-- a. acid i. 0.2000 p. 0.01000 d. NaOH j. 1.000 k. 0.005000 q. 0.001000 mol Mg(OH)2 We now can calculate the concentration of HCI since we have both the moles of HCI and the volume of HCI mol HCI b. base c. HCI )= LHCI r. 58.31 ) = e. NaCl I. 100.0 f. H₂O m. 0.1000 -) = s. 5.831 g. 25.00 mol Mg(OH)2 mol/L n. 0.05000 h. 0.02500 mol HCI o. 2.000

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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100.0 mL of HCI required 5.831 g Mg(OH)2 to titrate. The molar mass of Mg(OH)2 is 58.31 g. What is the concentration of the HCI ? The neutralization equation is 2 HCI + Mg(OH)2 MgCl₂ + 2 H₂O There is enough information to calculate the moles of Mg(OH)2 mol Mg(OH)2 -): (5.831 g Mg(OH)₂ (- g Mg(OH)₂ Then use stoichiometry to convert moles of Mg(OH)2 into moles of HCI mol HCI concentration a. acid i. 0.2000 mol Mg(OH)2 (-- mol Mg(OH)2 We now can calculate the concentration of HCI since we have both the moles of HCI and the volume of HCI mol HCI p. 0.01000 b. base c. HCI NaOH j. 1.000 k. 0.005000 q. 0.001000 L HCI r. 58.31 e. NaCl I. 100.0 f. H₂O m. 0.1000 s. 5.831 g. 25.00 mol Mg(OH)2 mol/L n. 0.05000 h. 0.02500 mol HCI o. 2.000
Transcribed Image Text:100.0 mL of HCI required 5.831 g Mg(OH)2 to titrate. The molar mass of Mg(OH)2 is 58.31 g. What is the concentration of the HCI ? The neutralization equation is 2 HCI + Mg(OH)2 MgCl₂ + 2 H₂O There is enough information to calculate the moles of Mg(OH)2 mol Mg(OH)2 -): (5.831 g Mg(OH)₂ (- g Mg(OH)₂ Then use stoichiometry to convert moles of Mg(OH)2 into moles of HCI mol HCI concentration a. acid i. 0.2000 mol Mg(OH)2 (-- mol Mg(OH)2 We now can calculate the concentration of HCI since we have both the moles of HCI and the volume of HCI mol HCI p. 0.01000 b. base c. HCI NaOH j. 1.000 k. 0.005000 q. 0.001000 L HCI r. 58.31 e. NaCl I. 100.0 f. H₂O m. 0.1000 s. 5.831 g. 25.00 mol Mg(OH)2 mol/L n. 0.05000 h. 0.02500 mol HCI o. 2.000
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