Multiple Choices: Choose the best (one) answer. Show in bold. Questions break-down: Chapter 8: Q1-8; Chapter 9: Q9-16: Chapter 10:

HCCS CHEM 1405 textbook PRACTICE EXAM III (Ch. 8-10) 5 th and 6 th edition of Corwin s The contents of these chapters are more calculation-oriented and are the beginning of learning of the chemical language. More study is required to comprehend the concepts. Probably the most difficult test among the four. Multiple Choices: Choose the best (one) answer. Show in bold. Questions break-down: Chapter 8: Q1-8; Chapter 9: Q9-16: Chapter 10: 17-25. Multiple Choices: Choose the best answer. The correct answer is shown in bold character. 1. Which of the following is evidence for a chemical reaction? (A) A gas is detected. (B) A precipitate or solid is formed. (C) A color change is observed. (D) An energy change is noted. (E) All of the above. Hint: See both 5 th ed. and 6 th ed. Section 8.1 2. Which of the following statement best describes the reaction below? Pb(NO 3 ) 2(aq) + K 2 CrO 4(aq) PbCrO 4(s) + 2KNO 3(aq) (A) Lead nitrate reacts with potassium chromate to give lead chromate and potassium nitrate. (B) Lead (II) nitrate reacts with potassium chromate to give lead (II) chromate and potassium nitrate. (C) Lead nitrate solution reacts with potassium chromate solution to give solid lead chromate and potassium nitrate solution. (D) Lead (II) nitrate aqueous solution reacts with potassium chromate aqueous solution to give solid lead (II) chromate and potassium nitrate aqueous solution. Hint: See both 5 th ed. and 6 th ed. Section 8.2. Also transition metal, lead, requires specify its charge as described in Sections 7.4 and 7.6. 3. If the equation C 4 H 10 + O 2 CO 2 + H 2 O is balanced, which of the following quantities is correct? Note: Coefficients must be integers. (A) 2 C 4 H 10, 13 O 2, 8 CO 2, and 10 H 2 O (B) 13 C 4 H 10, 2 O 2, 10 CO 2, and 8 H 2 O (C) C 4 H 10, 6.5 O 2, 4 CO 2, and 5 H 2 O (D) C 4 H 10, 2 O 2, 5 CO 2, and H 2 O Hint: See both 5 th ed. and 6 th ed. Section 8.3. However, you must read Section 8.2 first to be able to translate a chemical reaction into a chemical equation 1

when the question is given as a description. For instance, the above question can be given as Balance the equation when butane gas reacts with oxygen gas (combustion or burning) or write a balanced chemical equation when butane reacts with oxygen. Balancing equation is a very, very important question. To balance a chemical equation, you must make sure the number of atoms of each kind at both sides of the arrow is identical. Also start examining the most bulky species, that is, the one with the most different kinds of atoms and number of atoms. In this question, C 4 H 10 is the most bulky one. So we use it as reference by putting 1 in front of it to remind us we have done examining C 4 H 10. Now the equation is updated to be 1 C 4 H 10 + O 2 CO 2 + H 2 O Since C 4 H 10 contains 4 carbon atoms, so we need four carbon atoms at the right side, which leads us to put 4 (called coefficient) in front of the CO 2. Now the equation is updated to be 1 C 4 H 10 + O 2 4 CO 2 + H 2 O As there are 10 hydrogen atoms in C 4 H 10, thus we need to balance the hydrogen atoms, which lead us to put 5 in front of the H 2 O. Now the equation is updated to be 1 C 4 H 10 + O 2 4 CO 2 + 5 H 2 O Now we need to balance the oxygen atoms. Since there are 4x2+5x1 = 13 oxygen atoms at the right side, and thus the left side must have the same number. That is x2 = 13. So = 6.5 According to the rule, all the coefficients must be whole numbers and you cannot round them, so we must multiply 6.5 by a whole number, which must be 2 or greater, until it reaches an integer. We find that 2 will do the job. So each coefficient must be multiplied by 2 as well. That is, 2 x (1 C 4 H 10 + 6.5 O 2 4 CO 2 + 5 H 2 O) which leads to 2 C 4 H 10 + 13 O 2 8 CO 2 + 10 H 2 O Thus, we have balanced this equation. 4. Which of the following statement is wrong? (A) AgNO 3 3 (aq) is a precipitation reaction. (B) NH 4 3 (g) + HCl (g) is a decomposition reaction. (C) 2C 4 H 10 + 13O 2 8CO 2 + 10H 2 O is a combustion reaction. (D) HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) is a neutralization reaction. E) Zn(s) + 2 AgNO 3 (aq) 2 Ag(s) + Zn(NO 3 ) 2 (aq) is a double-replacement reaction 2

Hint: See both 5 th ed. and 6 th ed. Section 8.4 for summary and Sections 8.5 and 8.6 for examples. Please note that Choice (A) is also called a double-replacement reaction or a metathesis reaction. 5. What is the product predicted from the following combination reaction? Al(s) + O 2 (g) (A) AlO (B) AlO 2 (C) Al 2 O 3 (D) Al 3 O 2 (E) Al 2 O Hint: See both 5 th ed. and 6 th ed. Section 8.5. Please note that reaction with oxygen gas is also called combustion reaction or oxidation-reduction (i.e. redox) reaction. What are the products from the following decomposition reaction? CaCO 3 (s) (A) CaO (s) and CO 2 (g) (B) Ca(s) and CO 3 (g) (C) CaO 2 (g) and CO(g) (D) CaCO and O 2 (g) (E) 2 1 Hint: See both 5 th ed. and 6 th ed. Section 8.6. Please note that the decomposition reaction is the reverse reaction of combination. 6. Which of the following metals reacts with aqueous FeSO 4? Partial Activity Series: Mg > Al > Zn > Fe > (H) (A) Al (B) Mg (C) Zn (D) All of the above. (E) None of the above. Hint: See both 5 th ed. and 6 th ed. Section 8.7. Information (activity series for metals) at the margin of p. 214, 6 th ed. is very useful. The reaction direction in p. 214 is always from left to right. 7. What are the products from the following single-replacement reaction? Na(s) + H 2 O(l) 3

(A) Na 2 O and H 2 (B) Na 2 O and H 2 O (C) NaOH and H 2 (D) NaOH and H 2 O (E) No reaction. Hint: See both 5 th ed. and 6 th ed. Section 8.8. Watch out the charges of ions before combining them into an electrically neutral compound or element. What are the products from the following double-replacement reaction? Pb(NO 3 ) 2 + K 2 CrO 4 (A) Pb(NO 3 ) 2 and K 2 CrO 4 (B) K 2 CrO 4 and PbCrO 4 (C) PbCrO 4 and KNO 3 (D) Pb(NO 3 ) 2 and KNO 3 (E) All of the above. Hint: See both 5 th ed. and 6 th ed. Section 8.10. Watch out the charges of ions before combining them into an electrically neutral compound or element. 8. Which of the following solid compounds is insoluble in water? (A) Ba(OH) 2 (B) AgNO 3 (C) PbCl 2 (D) (NH 4 ) 2 CO 3 (E) Ca(NO 3 ) 2 Hint: See both 5 th ed. and 6 th ed. Section 8.9. Know how to use Table 8.2, p. 219 sixth ed. This is a very important question. Refer to the Solubility Rules in Appendix E in the 5 th ed. and Appendix D, p. 600, in the 6 th ed. of the textbook. Note that Ba(OH) 2 is soluble in water. 9. Which of the following statement is correct? (A) 1 mole of Ca atoms contains 6.02 x 10 23 Ca atoms. (B) 3.01 x 10 23 N 2 molecules is equivalent to 0.5 mole of N 2 molecules. (C) The Avogadro s number is 6.02 x 10 23. (D) There are 6.022 x 10 23 NaCl formula units in 1 mole of NaCl. (E) All of the above. Hint: See both 5 th ed. and 6 th ed. Section 9.1 for Avogadro s number and Section 9.2 for mole calculation. 10. What is the approximate molar mass of calcium nitrate, Ca(NO 3 ) 2? (A) 164.1 g/mol (B) 106.4 g/mol (C) 52 g/mol 4

(D) 98.5 g/mol Hint: See both 5 th ed. and 6 th ed. Section 9.3. In the formula, Ca(NO 3 ) 2, it indicates there are 1 Ca, 1x2 = 2 N and 3x2 = 6 O. Go to the periodic table and locate the atomic mass for each atom. Thus the formula mass of Ca(NO 3 ) 2 = 1x40 + 2x14 + 6x16 = 164. Note that if the unit given as amu (referring to atom mass unit it refers to one Ca(NO 3 ) 2; if the unit given as grams or g/mol it refers to contains 6.02 x 10 23 Ca(NO 3 ) 2 which is commonly called molar mass as 1 mole of substance contains Avogadro number of particles (atoms or molecules) or units. 11. A gas has a density of 3.84 g/l at a pressure of 3.50 atm and a temperature of 37 o C. What is the molar mass (g/mol) of this gas? (A) 27.89 (B) 0.102 (C) 497.28 (D) 58.73 Hint: See both 5 th ed. and 6 th ed. Sections 9.5 and 9.6. Be very careful that the method of Example Exercise 9.9 works only for 1 L gas at STP (0 o C and 1 atm) condition. This question the gas is at 3.5 atm and 37 o C and thus it does not apply. The easiest way to solve this question not under the standard condition is to apply the derived ideal gas equation: PM = drt. Here, P is pressure in atm, M is molar mass in gram/mol, d is density in gram/liter, R is the ideal gas constant, 0.082 atm.l/mol.k, T is temperature in Kelvin. Recall K = o C + 273.15. This equation, PM = drt, is derived from the ideal gas equation PV = nrt where P is pressure in atm, V is volume in leter, n is mole, R is the ideal gas constant, 0.082 atm.l/mol.k, and T is temperature in Kelvin. Since n = mole = mass in gram /molar mass, we can re-write PV = nrt to PV= (m/m)rt where m is the mass in gram and M is the molar mass in gram/mole. By switching the positions of V and M, we can re-write the equation to PM = (m/v)rt = drt where d is the density in gram/liter. So 3.50 x M = 3.84 x 0.082 x (37 + 273.15) = 97.66 M = 97.66/3.50 = 27.90 g/mole 12. What is the mass of 0.500 Liter of oxygen gas, O 2, at STP? (A) 0.286 g (B) 0.714 g 5

(C) 3.50 g (D) 6.40 g (E) 112 g Hint: See both 5 th ed. and 6 th ed. Sections 9.5 and 9.6. The method shown in the textbook Example Exercise 9.10 is (0.500/22.4) x 32 = 0.714 g as at STP one mole of oxygen gas weighs 32 grams and occupies 22.4 liters. STP refers to standard temperature and pressure, that is, 0 o C and 1 atm. The molar mass of O 2 is 32.0 g/mol. There are two other ways to solve this question: one is using PM = drt and another is applying PV=nRT and n=mass/molar mass. Using PM=dRT method: 1 x 32 = d x 0.082 x (0+273.15) d = 32/ 22.3983 = 1.43 g/l. So the mass of oxygen gas = 1.43 x 0.500 = 0.714 g. Applying PV=nRT and n=mass/molar mass method: 1 x 0.5 = n x 0.082 x 273.15 n = 0.5/22.3983 = 0.00223 mole Mass = mole x molar mass = 0.00223 x 32 = 0.714 gram 13. What is the percentage of oxygen in carbon dioxide, CO 2? (A) 12.37% (B) 27.27% (C) 36.36% (D) 57.14% (E) 72.73% Hint: See both 5 th ed. and 6 th ed. Section 9.7 Example Exercise 9.12. O% = 2 x {16.00/(1x12+2x16)} x 100% = {32/44}x100% = 72.73%. % composition of an atom = n x (molar mass of an atom/molar mass of compound) x 100% 14. If 0.250 mol V reacts with 0.375 mol O, what is the empirical formula of vanadium oxide? (A) VO (B) V 2 O 3 (C) V 2 O 5 (D) V 3 O 2 (E) V 5 O 2 Hint: See both 5 th ed. and 6 th ed. Section 9.8. Empirical formula must be obtained from experiment and it is simplest whole number mole ratio of atoms. V: O = 0.250 : 0.375 = (0.250/0.250) : (0.375/0.250) = 1 : 1.5 = 1x2 : 1.5x2 = 2 : 3 6

Do not round the numbers for no valid reasons. Note that 0.5 = 1/2; 0.33 or 0.34 = 1/3; 0.25 = ¼. 15. What is the empirical formula for methyl benzoate, a compound used in the manufacture of perfumes, contains 70.57% carbon, 5.94% hydrogen, and 23.49% oxygen? Note: subscripts must be integers. A) C 4 H 4 O B) C 2 H 2 O 0.5 C) C 8 H 8 O 2 D) CHO Hint: Section 9.8 Example Exercise 9.14 (sixth edition). This is a very, very important question. The empirical formula is the simplest integral ratio of moles among each atom. Here, there are three different kinds of atoms, C, H and O. Thus mole of C = 70.57/12 = 5.88; mole of H = 5.94/1 = 5.94; mole of O = 23.49/16 = 1.47. Note that as long as one of the moles is not an integer, we have to divide the smallest value among them: here the smallest value is 1.47. So C : H : O = 5.88/1.47 : 5.94/1.47 : 1.47/1.47 = 4 : 4.04 : 1. Since 4.04 is very close to 4.00 and thus we can round it to 4.00. So C : H : O = 4 : 4 : 1, which indicates that the empirical formula contains four C, four H and one O. Thus, the empirical formula is written as C 4 H 4 O as 1 is usually not written in the formula. *******There is a type of question that requires two steps to solve it. See the following vitamin C question.******* The composition of vitamin C or ascorbic acid is made by 40.92% C, 4.58% H and 54.50% O. What is the empirical formula for vitamin C? C : H : O = 40.92/12.01 : 4.58/1.01 : 54.50/16.00 = 3.407 : 4.54 : 3.406 = 1: 1.33 : 1 = 3 : 4 : 3. As long as one of the values is not an integer, we need to modify them by multiplying a whole number 2 or greater than 3 until all the numbers turn to be whole numbers. Since multiply by two will not make 2.66 as an whole number, and thus we try to multiply by three. This time it works. Please note that we can NOT round 3.407 or 3.406 to 3 or to 4 as it will cause too much error. Same as for 4.54 can NOT be rounded to 4 or 5. 16. The empirical formula of ethylene glycol is CH 3 O. The weight of 1 mole of ethylene glycol is 62.1 g. What is its molecular formula? Note: subscripts must be integers. A) C 2 H 6 O 2 B) CHO C) CH 3 O D) C 0.5 H 0.5 O 0.5 Hint: See both 5 th ed. and 6 th ed. Section 9.9 Example Exercise 9.15 (sixth edition). Remember the definition that the molecular formula is an integral multiple of empirical formula. That is, the molar mass (i.e. molecular weight) = 7

empirical formula weight x integer. So the empirical formula weight of CH 3 O = 12x1+1x3+16x1=31. So the integer = 62.1/31 = 2. Thus there are two empirical formulas in a molecular formula. Therefore, the molecular formula is C 2 H 6 O 2. 17. How many moles of oxygen gas, O 2, react with 2 moles of nitrogen monoxide gas, NO, according to the following equation? NO(g) + O 2 (g) NO 2 (g) (A) 1 mol. (B) 2 mol. (C) 3 mol. (D) 4 mol. (E) None of the above. Hint: See both 5 th ed. and 6 th ed. Section 10.1 in-text example. However, to be more accurate, this question involves concepts from Sections 10.1 to 10.3: balancing equation and stoichiometry, which deals with mole mole relationship. As the coefficients (the numbers proceed chemical formulas; shown in blanks to be balanced in the above question) represent the mole mole relationship. 18. How many moles of water react with 0.500 mol of lithium metal according to the following reaction? Li(s) + H 2 O LiOH(aq) + H 2 (g) (A) 0.125 mol. (B) 0.250 mol (C) 0.500 mol. (D) 1.00 mol (E) 2.00 mol. Hint: See both both 5 th ed. and 6 th ed. Section 10.2. First, balance the equation. Then apply mole-mole relationship according to the balance equation. Balanced equation: _2_ Li(s) + _2_ H 2 O _2_ LiOH(aq) + _1_ H 2 (g) Since the ratio for Li and H 2 O is 2 : 2 = 1 : 1 ratio, so the mole of water required will be the same mole number of Li consumed. 18. How many moles of water must react in order to produce 0.500 mol of potassium hydroxide? K(s) + H 2 O(l) KOH(aq) + H 2 (g) 8

(A) 0.125 mol. (B) 0.250 mol (C) 0.500 mol. (D) 1.00 mol (E) 2.00 mol. Hint: See both 5 th ed. and 6 th ed. Section 10.2. Same concept as Q 17. First, balance the equation. Then apply mole-mole relationship according to the balance equation. _2_ K(s) + _2_ H 2 O(l) _2_ KOH(aq) + _1_ H 2 (g) Since the ratio for KOH and H 2 O is 2 : 2 = 1 : 1 ratio, so the mole of water consumed will be the same mole number of KOH produced. 20. How many grams of aluminum metal must react to give 500.0 g of iron? FeO(l) + Al(l) Fe(l) + Al 2 O 3 (A) 80.6 g (B) 161 g (C) 242 g (D) 362 g (E) 483 g Hint: See both 5 th ed. and 6 th ed. Section 10.4: Mass-Mass problems. First, balance the equation. Then apply mole-mole relationship according to the balance equation. Then convert mole to mass. _3_ FeO(l) + _2_ Al(l) _3_ Fe(l) + _1_ Al 2 O 3 Mole of Fe = (500.0/55.845) = 8.953 mol {2 Al/ 3 Fe} = {y mol Al/8.953 mol Fe} => y = 2x8.953/3 = 5.969 mol Al Grams Al = 5.969 mol Al x 26.982 g/mol = 161.05 g Al Remember that there is no direct mass relationship in chemical reactions. See p. 284 (sixth edition) Concept Map Summary of Stoichiometry or the Road Map given by the instructor. 21. How many grams of mercuric oxide (216.59 g/mol) must decompose to release 0.375 L of oxygen gas at STP? HgO(s) Hg(l) + O 2 (g) 9

(A) 0.0752 g (B) 1.82 g (C) 3.63 g (D) 7.25 g (E) 14.5 g Hint:See both 5 th ed. and 6 th ed. Section 10.5: Mass-Volume problems. First, balance the equation. Then apply PV = nrt to calculate mole of oxygen. STP means 1 atm and 0 o C. Then apply mole relationship to calculate mol of HgO. Then cover mol of HgO to mass of HgO. _2_HgO(s) _2_ Hg(l) + _1_ O 2 (g) 1 x 0.375 = n x 0.082 x (0 + 273.15) n = 0.375/22.3983 = 0.01674 mol O 2 2 HgO/ 1 O 2 = y mol HgO / 0.01674 mol O 2 y = 2 x 0.01674 = 0.03348 mol HgO Grams of HgO = 0.03348 mol HgO x 216.59 gram/mol = 7.2514 grams HgO 22. Give the following equation, 3 BaCl 2 + 2 Na 3 PO 4 Ba 3 (PO 4 ) 2 + 6 NaCl, how many moles of Na 3 PO 4 will react with 0.45 mole of BaCl 2? (A) 0.3 (B) 13.3 (C) 3.5 (D) 4.6 (E) 2.4 Hint: See 5 th ed. Section 10.5 and 6 th ed. Section 10.2: Mole-Mole problems. From the equation, the involving species, 3 BaCl 2 and 2 Na 3 PO 4 with coefficients 3 and 2 respectively, tell us that for 3 moles of BaCl 2 it requires 2 moles of Na 3 PO 4. According to this proportion or ratio, 0.45 mole of BaCl 2 requires 0.45 x 2/3 = 0.3 mole of Na 3 PO 4. 23. If 1.00 mol of chromium reacts with 1.00 mol of oxygen gas according to the following reaction, how many moles of chromium (III) oxide are produced? Cr(s) + O 2 (g) Cr 2 O 3 (s) (A) 0.500 mol (B) 0.667 mol (C) 1.00 mol (D) 1.50 mol (E) 2.00 mol Hint: See both 5 th ed. and 6 th ed. Sections 10.7 and 10.8 Example Exercise 10.9: The limiting reactant concept. It s very, very important. First, balance the equation. Second apply the stoichiometric (that is mole-mole relationship according to the balanced equation). _4_Cr(s) + _3_ O 2 (g) _2_ Cr 2 O 3 (s) 10

If oxygen is the limiting reactant, then {3 O 2 / 2 Cr 2 O 3 } = {1 mol O 2 / y mol Cr 2 O 3 } y = 0.667 mol Cr 2 O 3 produced. If Cr is the limiting reactant, then {4 Cr/ 2 Cr 2 O 3 } = {1 mol Cr / z mol Cr 2 O 3 } z = 0.500 mol Cr 2 O 3 produced. Since 0.500 mol < 0.667 mol, thus there will be 0.500 mol Cr 2 O 3 actually produced and Cr is the limiting reactant and oxygen is the excess reactant. 24. If 82.4 g of aluminum metal (M=26.98 g/mol) reacts with 117.65 g of oxygen gas (M=32.00 g/mol) according to the following equation, how many grams of aluminum oxide (M=101.96 g/mol) are produced?. 4 Al(s) + 3 O 2 (g) 2 Al 2 O 3 (s) (A) 155.7 g (B) 250.0 g (C) 308.0 g (D) 350.4 g Hint: See both 5 th ed. and 6 th ed. Section 10.8 Example Exercise 10.10: Limiting reactant concept. Mole of Al = {82.4/26.98} = 3.054 mol Mole of oxygen gas = {117.65/32.00} = 3.677 mol *If Al is the limiting reactant, then 4 Al /2 Al 2 O 3 = 3.054 mol Al / y mole Al 2 O 3 y = (2x3.054)/4 = 1.527 mol Al 2 O 3 *If oxygen gas is the limiting reactant, then 3 O 2 /2 Al 2 O 3 = 3.677 mol O 2 / z mole Al 2 O 3 y = (2x3.677)/3 = 2.451 mol Al 2 O 3 Since 1.527 mol Al 2 O 3 < 2.451 mol Al 2 O 3, there will be actually 1.527 mol Al 2 O 3 produced and Al is the true limiting reactant. Grams of Al 2 O 3 = 1.527 mol x 101.96 g/mol = 155.7 g. This is called the theoretical yield in Section 10.9 because it is calculated from a balanced chemical equation. The actual yield must be physically weighed by the scale or balance. 11

25. Starting with 0.657 g of lead (II) nitrate, a student collects 0.905 g of precipitate. If the calculated mass of precipitate is 0.914 g (i.e. theoretical yield), what is the percent yield? (A) 71.9% (B) 72.6% (C) 99.0% (D) 101% (E) 138% Hint: See both 5 th ed. and 6 th ed. Section 10.9. Actual yield is the one that must be physically weighed. Percent yield = (0.905 g/ 0.914 g) x 100% = 99.02% Note that there is no unit in the percent yield. 12