STOICHIOMETRY REVIEW ANSWERS 1 The calculation of quantities

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STOICHIOMETRY REVIEW ANSWERS

STOICHIOMETRY REVIEW ANSWERS

 • 1. The calculation of quantities in chemical equations is called • Stoichiometry

• 1. The calculation of quantities in chemical equations is called • Stoichiometry • 2. The first step in most stoichiometry problems is to convert given quantities to • Moles • 3. When two substances react to form products, the reactant which is used up is called the • Limiting Reagent

 • 4. The amount of product that forms when a reaction is carried

• 4. The amount of product that forms when a reaction is carried out in the laboratory is the • Actual Yield • 5. The maximum amount of products that could be formed from given amounts of reactants is the • Theoretical Yield • 6. When an equation is used to calculate the amount of product that will form during a reaction, then the value obtained is called the • Theoretical Yield

 • 7. 2 Al(s) + 3 Pb(NO 3)2(aq) 2 Al(NO 3)3(aq) + 3

• 7. 2 Al(s) + 3 Pb(NO 3)2(aq) 2 Al(NO 3)3(aq) + 3 Pb(s) • A. ) 2 moles Al + 3 molecules Pb(NO 3)2 2 molecules Al(NO 3)3 + 3 moles Pb • F • • B. ) 2 grams Al + 3 grams Pb(NO 3)2 2 grams Al(NO 3)3 + 3 grams Pb F • • C. ) 2 moles Al + 3 moles Pb(NO 3)2 2 moles Al(NO 3)3 + 3 moles Pb T • 8. In every chemical reaction ______ stay the same. • The Mass • 9. In the reaction 2 CO + O 2 2 CO 2, what is the ratio of moles of oxygen used to moles of CO 2 produced? • 1: 2

 • 10. For C 5 H 12 + 8 O 2 5 CO

• 10. For C 5 H 12 + 8 O 2 5 CO 2 + 6 H 2 O • ______ moles of reactants chemically change into ______ moles of product. • 9 moles of reactant to 11 moles of product

11. How many moles of oxygen are required to react completely with 14. 8

11. How many moles of oxygen are required to react completely with 14. 8 mol Al when given 4 Al + 3 O 2 2 Al 2 O 3? 14. 8 -----mol Al x 3 mol O 2 --------4 mol ----Al = 11. 1 mol O 2

12. Given the balanced equation 16 HCl + 2 KMn. O 4 2 KCl

12. Given the balanced equation 16 HCl + 2 KMn. O 4 2 KCl + 2 Mn. Cl 2 + 5 Cl 2 + 8 H 2 O if 2. 0 mol of KMn. O 4 reacts, how many moles of H 2 O are produced? 8. 0 mol H 2 O

13. Based on… 16 HCl + 2 KMn. O 4 2 KCl + 2

13. Based on… 16 HCl + 2 KMn. O 4 2 KCl + 2 Mn. Cl 2 + 5 Cl 2 + 8 H 2 O …how many grams of KCl are produced when 3. 0 mol of Kmn. O 4 reacts? ------3. 0 mol KMn. O 4 2 mol ----KCl 74. 5 g KCl x -------- X ----------2 mol ------KMn. O 4 1 mol ----KCl = 220 g KCl

1. 14. If Cu. O + H 2 Cu + H 2 O, how

1. 14. If Cu. O + H 2 Cu + H 2 O, how many moles of H 2 O are produced when 240 grams of Cu. O react? 1 mol ----Cu. O 1 mol H 2 O ----240 g Cu. O x -------- X ----------80 g----Cu. O 1 mol ----Cu. O = 3. 0 mol H 2 O

15. In the reaction Zn + H 2 SO 4 Zn. SO 4 +

15. In the reaction Zn + H 2 SO 4 Zn. SO 4 + H 2 how many grams of H 2 SO 4 are required to produce 2. 3 gram of H 2? 2. 3 g------H 2 x 1 mol H 2 ----------------- 2 g------H 2 1 mol H 2 SO 4 -------- 98 g H 2 SO 4 -----1 mol H 2 SO 4 ----------------- x ----------- 113 g H 2 SO 4 =

16. Li. OH + HBr Li. Br + H 2 O If you start

16. Li. OH + HBr Li. Br + H 2 O If you start with 15. 0 grams of lithium hydroxide, how many grams of lithium bromide will be produced? 54. 4 g Li. Br

17. How many molecules of oxygen are required to react completely with 2. 96

17. How many molecules of oxygen are required to react completely with 2. 96 X 1023 molecules Al when given 4 Al + 3 O 2 2 Al 2 O 3? 3 molecules O 2 -----------2. 96 x 1023 molecules Al x --------- 4 -----------molecules Al 2. 22 x 23 10 = molecules O 2

18. How many molecules of oxygen are produced by the decomposition of 6. 54

18. How many molecules of oxygen are produced by the decomposition of 6. 54 g of potassium chlorate (KCl. O 3)? 2 KCl. O 3 2 KCl + 3 O 2 ------1 mol KCl. O 3 ----3 mol O 2 122. 5 g------KCl. O 3 ------2 mol KCl. O 3 6. 02 x 1023 molecules O 2 6. 54 g------KCl. O 3 x --------------- x --------- 4. 82 x 22 10 ----1 mol O 2 molecules O 2 =

 • 19. Li. OH + KCl Li. Cl + KOH • A) I

• 19. Li. OH + KCl Li. Cl + KOH • A) I began this reaction with 20 grams of lithium hydroxide. What is my theoretical yield of lithium chloride? Convert 20 grams of Li. OH Grams of Li. Cl 35 g Li. Cl • B) I actually produced 6 grams of lithium chloride. What is my percent yield? Actual _____ Theoretical X 100 = 17%

 • 20. C 3 H 8 + 5 O 2 3 CO 2

• 20. C 3 H 8 + 5 O 2 3 CO 2 + 4 H 2 O • A) If I start with 5 grams of C 3 H 8, what is my theoretical yield of water? Convert 5 grams C 3 H 8 Grams of H 2 O 8. 2 g H 2 O • B) I made 6. 1 g of water. What was my percent yield? Actual _____ Theoretical X 100 = 74%

21. Be + 2 HCl Be. Cl 2 + H 2 My theoretical yield

21. Be + 2 HCl Be. Cl 2 + H 2 My theoretical yield of beryllium chloride was 10. 7 grams. If my actual yield was 4. 5 grams, what was my percent yield? ___ Actual Theoretical X 100 = 42%

22. Ca. SO 4 + Na. I Ca. I 2 + Na 2 SO

22. Ca. SO 4 + Na. I Ca. I 2 + Na 2 SO 4 If 34. 7 g of calcium sulfate and 58. 3 g of sodium iodide are placed in a reaction vessel, what will be the limiting reagent? Na. I Note: Refer to Limiting Reagent Notes to see how to work it out.