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#Chem 161-2007 Chapter 3B-Stoichiometry practice problems

# #Chem 161-2007 Chapter 3B-Stoichiometry practice problems -...

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CHEM 161-2007 CHAPTER 3B - STOICHIOMETRY PRACTICE PROBLEMS DR. ED TAVSS Balancing chemical reactions Reaction stoichiometry Solution Conc.: Molarity Limit. reactants & % yield

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BALANCING CHEMICAL REACTIONS 10 Chem 161-2006 Exam I Hill, Petrucci et al., 4 th edition Chapter 3 – Stoichiometry: Chemical Calculations Balancing chemical equations In the combustion of hexane, C 6 H 14 , reacts with O 2 to produce carbon dioxide and water. How many moles of O 2 are required to react with 1.0 mole of hexane? A . 9.5 B. 19 C. 9 D. 9 E. 3 C 6 H 14 + O 2 → CO 2 + H 2 O C 6 H 14 + O 2 → 6CO 2 + H 2 O C 6 H 14 + O 2 → 6CO 2 + 7H 2 O C 6 H 14 + 19/2 O 2 → 6CO 2 + 7H 2 O 2C 6 H 14 + 19O 2 → 12CO 2 + 14H 2 O A
24 Chem 161-2005 Hourly Exam II Chapter 3 - Stoichiometry Balancing chemical equations In the combustion of butane, C 4 H 10 , butane reacts with O 2 to produce carbon dioxide and water. When the equation for this combustion is balanced with the smallest possible integer coefficients, what is the coefficient of O 2 A. 4 B. 5 C. 7 D. 9 E . 13 C 4 H 10 + O 2 → CO 2 + H 2 O C 4 H 10 + O 2 → 4CO 2 + H 2 O C 4 H 10 + O 2 → 4CO 2 + 5H 2 O C 4 H 10 + 13/2 O 2 → 4CO 2 + 5H 2 O 2C 4 H 10 + 13O 2 → 8CO 2 + 10H 2 O 14. Chem 161-2004 Exam I Zumdahl 6 th edition Chapter 3 Balancing Chemical Equations Consider the reaction wPCl 5 + xH 2 O yPOCl 3 + zHCl This equation is balanced when A. choose this choice if none of the others is correct B . w = 1, x = 1, y = 1, z = 2 C. w = 2, x = 2, y = 2, z = 1 D. w = 2, x = 2, y = 2, z = 2 E. w =1, x =2, y = 2, z = 4 wPCl 5 + xH 2 O yPOCl 3 + zHCl Begin with the most complex molecule, which is either PCl5 or POCl3. The P’s are already balanced. Balance the Cl’s. The coefficient to balance the Cl’s must go on the HCl because if the

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coefficient goes on the POCl3 then the P’s will no longer be balanced. PCl 5 + H 2 O POCl 3 + 2HCl Everything is now balanced. ZUMDAHL 5 TH EDITION CHEM 161-2002 RECITATION 4 TH WEEK CHAPTER 3 - STOICHIOMETRY BALANCING CHEMICAL REACTIONS 3-83 Balance the following equations representing combustion reactions: a. C 12 H 22 O 11(s) + O 2(g) → CO 2(g) + H 2 O (g) a. C 12 H 22 O 11(s) + 12O 2(g) → 12CO 2(g) + 11H 2 O (g) b. C 6 H 6(l) + O 2(g) → CO 2(g) + H 2 O (g) b. C 6 H 6(l) + (15/2)O 2(g) → 6CO 2(g) + 3H 2 O (g) b. 2C 6 H 6(l) + 15O 2(g) → 12CO 2(g) + 6H 2 O (g) c. 2Fe (s) + (3/2)O 2(g) → Fe 2 O 3(s) c. 4Fe (s) + 3O 2(g) → 2Fe 2 O 3(s) d. C 4 H 10(g) + O 2(g) → CO 2(g) + H 2 O (g) d. C 4 H 10(g) + (13/2)O 2(g) → 4CO 2(g) + 5H 2 O (g) d. 2C 4 H 10(g) + 13O 2(g) → 8CO 2(g) + 10H 2 O (g) e. FeO (s) + O 2(g) → Fe 2 O 3(s) e. 2FeO (s) + (1/2)O 2(g) → Fe 2 O 3(s) e. 4FeO (s) + O 2(g) → 2Fe 2 O 3(s)
CHEM 161-1999 FINAL EXAM + ANSWERS CHAPTER 3 - STOICHIOMETRY BALANCING CHEMICAL EQUATIONS

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