CHEM107_Exam2_Study Guide Answer

CHEM107_Exam2_Study - Texas A M University Department of Chemistry CHEM107 Spring 2011 Instructor Masud Monwar Ph.D Study Guide for Exam 2 28th

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Department of Chemistry CHEM107 Spring 2011 Instructor: Masud Monwar, Ph.D. Study Guide for Exam 2 28th February, 2011 Chapter 4: Stoichiometry 1. What mass of HCl, in grams, is required to react with 0.750 g of Al(OH) 3 . What mass of H 2 O, in grams, is produced? What mass of AlCl 3 , in grams, is produced? The equation for the reaction is: Al(OH) 3 + HCl ----------- AlCl 3 + H 2 O Solution: Molar mass of Al(OH) 3 = (1 × 27.0) + (3 × (1.0 + 16.0)) = 78.0 g/mol Molar mass of HCl = (1 × 1.0) + (1 × 35.45) = 36.45 g/mol Molar mass of H 2 O = (2 × 1.0) + (1 × 16.0) = 18.0 g/mol Molar mass of AlCl 3 = (1 × 27.0) + (3 × 35.45) = 133.35 g/mol 0.750 g of Al(OH) 3 × = 9.615 × 10 -3 mol of Al(OH) 3 9.615 × 10 -3 mol of Al(OH) 3 × = 0.0288 mol of HCl required 0.0288 mol of HCl × = 1.04976 g = 1.050 g of HCl required 9.615 × 10 -3 mol of Al(OH) 3 × = 9.615 × 10 -3 mol of AlCl 3 produced Mass of AlCl 3 produced = 9.615 × 10 -3 mol of AlCl 3 × = 1.282 g of AlCl 3 9.615 × 10 -3 mol of Al(OH) 3 × = 0.0288 mol of H 2 O produced Mass of H 2 O produced = 0.0288 mol of H 2 O × = 0.5184 g of H 2 O Continued to page 2 2. Consider the following reaction: P 4 (s) + 6Cl 2 (g) ------------ 4PCl 3 (s) Calculate the mass of phosphorus (P 4 ) required to produce 15.0 g of phosphorus trichloride, PCl 3 ?
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Solution: Molar mass of P 4 = (4 × 30.97) = 123.88 g/mol Molar mass of PCl 3 = (1 × 30.97) + (3 × 35.45) = 137.32 g/mol Mol of PCl 3 = 15.0 g of PCl 3 × = 0.1092 mol of PCl 3 Mols of P 4 required = 0.1092 mol of PCl 3 × = 0.0273 mol of P 4 required Mass of P 4 required = 0.0273 mol of P 4 × = 3.382 g of P 4 3. a. What is the molarity of the solution when 2.0 g of KOH is dissolved in 500 mL of solution? Solution: Molar mass of KOH = (39.0 + 16.0 + 1.0) = 56.0 g/mol 2.0 g KOH × = 0.0357 mol 500 mL × = 0.5 L Molarity = = = 0.071 mol/L or, M b. What is the mass of solute, in grams, in 250.0 mL of a 0.0125 M solution KMnO 4 ? Solution: Molar mass of KMnO 4 = (39.0 + 54.94 + 4×16.0) = 157.4 g/mol 250 mL × = 0.25 L Molarity = 0.0125 mol/L = = 0.071 mol/L or, M Number of mols = 0.0125 mol/L × 0.25 L = 3.125 × 10 -3 mol Mass of KMnO 4 = 3.125 × 10 -3 mol × = 0.49 g Continued to page 3 c. What volume of 2.06 M KMnO 4 , in liters, contains 322.0 g of solute? Solution: Molar mass of KMnO 4 = (39.0 + 54.94 + 4×16.0) = 157.4 g/mol 322.0 g of KMnO 4 × = 2.045 mol Molarity = 2.06 mol/L = Volume(L) = = 0.9927 L × = 992.7 mL
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d. If 4.00 mL of 0.0250 M CuSO 4 is diluted to 10.0 mL with pure water, what is the molar concentration of copper(II) sulfate in the diluted solution? Solution: M1V1 = M2V2 4.00 × 10 -3 L × 0.0250 mol/L = M2 × 10.0 × 10 -3 L M2 = = 0.01 mol/L or, M 4. What mass of HCl, in grams, is required to react with 0.750 g of Al(OH) 3 . What mass of H 2 O, in grams, is produced? What mass of AlCl 3 , in grams, is produced? The equation for the reaction is:
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This note was uploaded on 11/05/2011 for the course CHEM 107 taught by Professor Generalchemforeng during the Spring '07 term at Texas A&M.

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CHEM107_Exam2_Study - Texas A M University Department of Chemistry CHEM107 Spring 2011 Instructor Masud Monwar Ph.D Study Guide for Exam 2 28th

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