STUDENT REVIEW PACKET ANSWER KEY

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Ira Copeland
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1 STUDENT REVIEW PACKET ANSWER KEY Molar Mass - Find the masses of the following. Show all work: 1) N 2 2(14.0) = 28.0 g/mole 2) H 2 O 2(1.0) + 1(16.0) = 18.0 g/mole 3) NaCl 1(23.0) + 1(35.5) = 58.5 g/mole 4) Na 3 PO 4 3(23.0) + 1(31.0) + 4(16.0) = g/mole 5) Au 1(197.0) = g/mole 6) (NH 4 ) 2 CO 3 2(14.0) +8(1.0) + 1(12.0) + 3(16.0) = 96.0 g/mole 7) C 6 H 12 O 6 3(12.0)+12(1.0)+6(16.0) = g/mol 8) Ca 3 (PO 4 ) 2 3(40.1)+2(31.0)+8(16.0) = 310.3g/mol 9) (NH 4 ) 2 SO 4 2(14.0)+8(1.0)+1(32.1)+4(16.0) = 132.1g/mol Percent Composition Determine the percent composition for each of the following. 1) SO 2 1(32.1) + 3(16.0) = 64.1 g (32.1/64.1)*100 % = 50.1 % S (32.0/64.1)*100 % = 49.9 % O 2) NaCl 1(23.0) + 1(35.5) = 58.5 g (23.0/58.5)*100 % = 39.3 % Na (35.5/58.5)*100 % = 60.7 % Cl 3) H 2 SO 4 2(1.0) + 1(32.1) + 4(16.0) = 98.1 (2(1.0)/98.1)*100 % = 2.04 % H (32.1/98.1)*100 % = % S (4(16.0)/98.1)*100 % = %O 4) K 2 S 2(39.1) + 1(32.1) = g (2(39.1)/110.3)*100 % = 70.9 % K (32.1/110.3)*100 % = 29.1 % S 5) C 6 H 12 O 6 6(12.0) + 12(1.0) + 6(16.0) = g (6(12.0)/180)*100 % = 40.0 % C (12(1.0)/180)*100 % = 6.7 % H (6(16.0)/180)*100 % = 53.3 % O Page 83

2 1-Step Mole Conversion Convert the following. Show all work. Be sure to show the correct number of significant figures and units in your responses g NaOH to moles NaOH 1(23.0) + 1(16.0) + 1(1.0) = 40.0 g/mole NaOH 25.0 g NaOH X 1 mole NaOH = mole NaOH g NaOH moles H 3 PO 4 to g H 3 PO 4 3(1.0) + 1(31.0) + 4(16.0) = 98.0 g/mole H 3 PO moles H 3 PO 4 X 98.0 g H 3 PO 4 = 66 g H 3 PO mole H 3 PO moles SO 2 to L SO moles SO 2 X 22.4 L SO 2 = 287 L SO mole SO moles CO 2 to L CO moles CO 2 X 22.4 L CO 2 = 0.15 L CO mole CO moles NH 3 to molecules NH moles NH 3 X 6.02 x molecules NH 3 = 7.40 x molecules NH mole NH x molecules KOH to moles KOH 6.8 x molecules KOH X 1 mole KOH = 1.1 x 10 4 mole KOH (11000) x molecules KOH g LiCl to moles LiCl 1(6.9) + 1(35.5) = 42.4 g 90.0 g LiCl X 1 mole LiCl = 2.12 mole LiCl g LiCl moles F 2 to atoms F mole F 2 x 6.02 x molecules F 2 X 2 atoms F = 2.0 x atoms F mole F 2 1 molecule F 2 If you just solved for molecules your answer would be 1.0 x molecules x atoms Cu to moles Cu 5.23 x atoms Cu X 1 mole Cu = 8.69 mole Cu x atoms Cu mole C 2 H 2 to molecules C 2 H 2 Page 84

3 22.4 mole C 2 H 2 X 6.02 x molecules C 2 H 2 = 1.35 x10 25 molecules C 2 H mole C 2 H 2 2-Step Mole Conversion Convert the following. Show all work. Be sure to show the correct number of significant figures and units in your responses. 1. How many molecules of H 2 O are in 50.0 g H 2 O? 2(1.0) + 1(16.0) = 18.0 g 50.0 g H 2 O X 1 mole H 2 O X 6.02 x molecules H 2 O = 1.67 x molecules H 2 O g H 2 O 1 mole H 2 O 2. Find the volume of CO 2 in g CO 2. 1(12.0) + 2(16.0) = 44.0 g g CO 2 X 1 mole CO 2 X 22.4 L CO 2 = g CO g CO 2 1 mole CO 2 3. A sample of diamond has a mass of 0.35 g. How many carbon atoms does it contain? (You may assume that the diamond is pure carbon) g C X 1 mole C X 6.02 x atoms C = 1.8 x atoms C g C 1 mole C 4. A typical helium tank contains L of He. How many atoms is this? L He X 1 mole He X 6.02 x atoms He = x atoms He L He 1 mole He 5. How many grams do 7.12 x molecules of K 2 CO 3 weigh? 2(39.1) + 1(12.0) + 3(16.0) = g 7.12 x molecules K 2 CO 3 X 1 mole K 2 CO 3 X g K 2 CO x molecules K 2 CO 3 1 mole K 2 CO 3 = 1.63x 10 7 g K 2 CO 3 Page 85

4 Part II 1. What information is derived from the coefficients in a balanced equation? The coefficients in a balanced equation tell the relative number of moles of reactants and products. 2. What is a mole ratio? Why is the mole ratio important in calculating stoichiometric relationships? Mole ratio a conversion factor which relates the moles of a given substance to the moles of the desired substance The mole ratio is used when solving stoichiometric problems to convert from one substance to another substance. One starts with the given quantity of one substance and multiplies it by the form of the mole ratio that allows the given unit to cancel. 3. Which of these is the mole ratio used to convert grams of oxygen to grams of water in Question #2 (above)? 1. 1 mole O g 2. 1 mole O 2 2 moles H 2 O 3. 2 moles H 2 O Answer 1 mole O g 1 mole H 2 O 4. Balance each equation. a) _4_ H 2 + Fe 3 O 4 _3_ Fe + _4_ H 2 O b) _2_ Al(OH) 3 Al 2 O 3 + _3_ H 2 O c) P 2 O 5 + _3_ H 2 O _2_ H 3 PO 4 d) _2_ Na + _2_ H 2 O _2_ NaOH + H 2 e) Pb(NO 3 ) 2 + _2_ NaI PbI 2 + _2_ NaNO 3 f) _3_ Ag 2 SO 4 + _2_AlCl 3 _6_ AgCl + Al 2 (SO 4 ) 3 g) _2_ C 2 H 2 + _5_ O 2 _4_ CO 2 + _2_ H 2 O Page 86

5 Use dimensional analysis or an acceptable Problem Solving Format. Be sure to include units, and report your answer to the correct number of significant figures. 5. Carbon disulfide is an important industrial solvent. It is prepared by the reaction of coke (carbon) with sulfur dioxide. 5C(s) + 2SO 2 (g) CS 2 (l) + 4CO(g) How many moles of carbon are needed to react with 5.44 moles of SO 2? 5.44 moles SO 2 x 5 moles C = 13.6 moles C 2 moles SO 2 How many moles of carbon monoxide form at the same time as moles of CS 2 form? moles CS 2 x 4 moles CO = moles CO 1 mole CS 2 5. How many moles of SO 2 are required to make 118 moles of CS 2? 118 moles CS 2 x 2 moles SO 2 = 236 moles SO 2 1 mole CS 2 6. Lithium nitride reacts with water to form ammonia and aqueous lithium hydroxide. Li 3 N(s) + 3H 2 O(l) NH 3 (g) + 3LiOH(aq) What mass of water is needed to react with 32.9 g Li 3 N? 32.9g Li 3 N x 1 mole Li 3 N x 3 moles H 2 O x g = 51.1g H 2 O g 1 mole Li 3 N 1 mole H 2 O When the above reaction takes place, how many molecules of NH 3 are produced? 32.9g Li 3 N x 1 mole Li 3 N x 1 mole NH 3 x 6.02 x RP = 5.69 x molecules g 1 mole Li 3 N 1 mole Page 87

6 7. The reaction of iron(iii) oxide with carbon monoxide produces iron and carbon dioxide. Fe 2 O 3 (s) + CO(g) Fe(s) + CO 2 (g) Balance the skeleton equation. Fe 2 O 3 (s) + 3 CO(g) 2 Fe(s) + 3 CO 2 (g) If you have 39.5 g Fe 2 O 3, how many grams of CO are required for complete reaction? 39.5g Fe 2 O 3 x 1 mole Fe 2 O 3 x 3 mole CO x g CO = 20.8g CO g 1 mole Fe 2 O 3 1 mole CO How many grams of Fe 2 O 3 are required to produce 4.65 g Fe? 4.65g Fe x 1 mole Fe x 1 mole Fe 2 O 3 x g = 6.65g Fe 2 O g 2 moles Fe 1 mole Fe 2 O 3 How many grams of iron can be produced from 145 g Fe 2 O 3? 145g Fe 2 O 3 x 1 mole Fe 2 O 3 x 2 moles Fe x g = 101g Fe g 1 mole Fe 2 O 3 1 mole Fe When 67.8 g Fe 2 O 3 reacts with an excess of CO, 44.1 g Fe is produced. What is the percent yield? Theoretical Yield 67.8g Fe 2 O 3 x 1 mole Fe 2 O 3 x 2 moles Fe x g = 47.4g Fe g 1 mole Fe 2 O 3 1 mole Fe Percent Yield % Yield = Experimental Yield x 100 Theoretical Yield = 44.1g Fe x 100 = 93.0 %Yield 47.4g Fe Page 88

7 8. The complete combustion of octane, C 8 H 18, in the presence of oxygen produces carbon dioxide and water. heat C 8 H 18 (g) + O 2 (g) CO 2 (g) + H 2 O(g) Balance the skeleton equation for the complete combustion of octane. heat 2 C 8 H 18 (g) + 25 O 2 (g) 16 CO 2 (g) + 18 H 2 O(g) What volume of CO 2 gas is released when 7.40 moles of octane react with excess oxygen at STP? 7.40 moles C 8 H 18 x 16 moles CO 2 x 22.4L = 1.33 x 10 3 L CO 2 2 moles C 8 H 18 1 mole What volume of oxygen gas is needed to react completely with 43.4 g octane? 43.4g C 8 H 18 x 1 mole C 8 H 18 x 25 moles O 2 x 22.4L = 106L O g 2 moles C 8 H 18 1 mole 9. Heating an ore of antimony, Sb 2 S 3, in the presence of excess iron ore gives the element antimony, Sb, and iron(ii) sulfide. heat Sb 2 S 3 (s) + 3Fe(s) 2Sb(s) + 3FeS(s) The following data was recorded when the reaction was carried out in the laboratory. DATA TABLE Constant mass of evaporating dish 45.50g Mass of evaporating dish and Sb 2 S g Mass of Sb 2 S g Mass of evaporating dish and Sb First heating 55.42g Mass of evaporating dish and Sb Second 55.34g heating Constant mass of evaporating dish and Sb 55.34g Mass of Sb (EY) 9.84g a) Complete the Data Table by filling in the mass of Sb 2 S 3 used in the reaction and the mass of Sb formed as the result of the reaction. b) Calculate the percent yield of antimony, Sb, in this reaction. Theoretical Yield 15.00g Sb 2 S 3 x 1 mole Sb 2 S 3 x 2 moles Sb x g = 10.75g Sb g 1 mole Sb 2 O 3 1 mole Sb Percent Yield Page 89 % Yield = EY x 100 = 9.84g Sb x 100 = 91.5 % Yield TY 10.75g Sb

8 10. List at least three (3) reasons why the percent yield of a product is usually less than 100%. reactions do not always go to completion impure reactants may give unwanted products competing side reactions give unwanted products loss of product during filtration or in transferring between containers 11. Describe a real world example of percent yield by applying the concept to a discipline other than chemistry. Be sure to include these terms in your answer: actual yield, predicted or theoretical yield, and percent yield. Answers may vary. A logical response might relate the concept of percent yield to agricultural crop yields or stock market gains. Page 90

11 Stoichiometry. Section 11.1 What is stoichiometry?

11 Stoichiometry. Section 11.1 What is stoichiometry? 11 Stoichiometry Section 11.1 What is stoichiometry? In your textbook, read about stoichiometry and the balanced equation. For each statement below, write true or false. 1.. 3. 4. 5. The study of the quantitative