{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# CHEM107_Exam2_Study Guide Answer - Texas A M University...

This preview shows pages 1–4. Sign up to view the full content.

Texas A & M University 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 ?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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: Al(OH) 3 + HCl ----------- AlCl 3 + H 2 O Solution: Balance the equation: Al(OH) 3 + 3HCl ----------- AlCl 3 + 3H 2 O Molar mass of Al(OH) 3 = 27.0 + (3×17.0) = 78.0 g/mol Molar mass of HCl = (1.0 + 35.45) = 36.45 g/mol Molar mass of H 2 O = (2 × 1.0 + 16.0) = 18.0 g/mol Molar mass of AlCl 3 = (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 HCl required 9.615 × 10 -3 mol of Al(OH) 3 × = 9.615 × 10 -3 mol of AlCl 3 produced 9.615 × 10 -3 mol of Al(OH) 3 × = 0.0288 mol of H 2 O produced Continued to page 4 5. Ammonia gas can be prepared by the reaction of a metal oxide such as calcium oxide with ammonium chloride.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}