Chem 1110 Chap. 3 WileyPlus

In 36 mg of water there are 2.0 x 10-3 molecules of water. False True

False Why: Can you have this number (2.0 x 10-3 or 0.0020) of molecules of anything? The answer is no! What is the smallest number of molecules you could have? It has to be 1! You can never have less than one molecule? So, does the number given have any meaning? Yes. Perhaps you did the following: 36 mg = 0.036 g and (0.036 g H2O) x (1 mol H2O / 18 g H2O) = 2.0 x 10-3 mol of H2O. The correct answer to this problem in units of molecules is 1.2 x 1021 molecules of water.

Consider the reaction below: A₁₄C₃ + 12 H₂O → 3 CH₄ + 4 Al(OH)₃. The yield of methane (CH4) from 2.0 mol of water is a) 0.50 mol b) 0.67 mol c) 4.0 mol d) 6.0 mol

a) 0.50 mol Why: Al4C3 + 12 H2O → 3 CH4 + 4 Al(OH)3 and determined that 3/12 or 0.25 mol of methane is produced per one mole of water. This would be 0.50 mole of methane for 2 moles of water. Using dimensional analysis (conversion factors from the equation) this can be shown as (2 mol H2O) X (3 mol CH4/ 12 mol H2O) = 0.50 mol CH4

Para-aminobenzoic acid, PABA, is an organic compound that has been used as a component in suntan lotions. The formula of PABA is C7H7NO2. Using the periodic table in your courseware and what you learned in the chapter on measurements, determine the molecular mass of PABA. a) 137.14 g/mol b) 151.15 g/mol c) 139.12 g/mol d) 173.14 g/mol

a) 137.14 g/mol

The mass of one atom of lead (Pb, atomic weight = 207.2 u) is a) 3.44 x 10-22 g b) 4.83 x 10-3 g c) 2.91 x 1021 g d) 8.02 x 10-27 g

a) 3.44 x 10-22 g Why: Since the atomic weight is numerically the same as the molar mass, this number in conjunction with Avogadro's number will give you the mass of one atom: (1 mol Pb / 6.02 x 1023 atoms Pb) x (207.2 g Pb / mol Pb) = 3.44 x 10 -22 g Pb / atom Pb.

If all three cost the same, which would be the cheapest source of chlorine? Atomic weights to use: Mg = 24.3 g/mol, Ca = 40.1 g/mol, O = 16.0 g/mol, Cl = 35.5 g/mol, and Tl = 204.3 g/mol. a) Ca(OCl)₂ b) Ca(OCl)₂ or Mg(ClO₂)₂ within 0.1% c) TlCl₃ d) Mg(ClO2)₂

a) Ca(OCl)₂ Why: Ca(OCl)₂ (40*1+2*16+2*35.5)/ 143.1 = 49.62% Cl Mg(ClO₂)₂ (24.3*1+2*35.5+4*16)/ 159.3 = 44.57% Cl TiCl₃ (204.3*1+3*35.5)/ 310.8 = 38.8% Cl

The empirical formula of a compound... a) can be used to calculate the percentage composition of the compound. b) is the only thing you need to find the molar mass of the compound. c) shows the structure of the molecule. d) has no relation to the molecular formula.

a) can be used to calculate the percentage composition of the compound.

The empirical formula of the compound with the following percent composition: 36.8% N and 63.2% O is a) NO₂ b) N₂O₃ c) N₃O₄ d) None of the above

b) N₂O₃ Why: The empirical (or simplest) formula gives the smallest whole number ratio of the atoms in a molecule. To determine the empirical formula you need the percentages of each element in the compound as well as the atomic weights of these elements. Start with the percent of each element. Percent means out of every 100 units - the units can be grams, but it is easier to use atomic mass units (u). Atomic weights for the elements are in u per atom (u/atom). Thus for N, [36.8 u / 14.0 u/atom] = 2.63 atoms N for O, [63.2 u / 16.0 u/atom] = 3.95 atoms O We therefore have 2.63 atoms N : 3.95 atoms O. The ratio of atoms of O to atoms of N is found by dividing both numbers by the smaller one. This is 1.50 atoms O per 1 atom N or 3 atoms of O to 2 atoms of N. The correct formula is N2O3.

A sample of boron (atomic symbol B; atomic weight = 10.8 u) has a mass of 0.550 g. How many moles of boron are there in this sample? a) 19.6 mol b) 0.0509 mol c) 5.94 mol

b) 0.0509 mol Why: The ability to convert moles to grams or the reverse is critical to understanding chemical reactions. Just as we would use the conversion factor 1 m = 100 cm (or 1 m / 100 cm) to convert a given number of centimeters to meters, a similar conversion factor for moles to grams can be used in chemical examples. The conversion factor is the molar mass of the substance under consideration. For an element the molar mass is simply the atomic mass expressed in units of grams per mole (or g/mol). For boron, the molar mass is 10.8 g/mol.A second feature in doing conversions is that all units except that wanted must cancel when doing the calculation. You don't just multiply or divide numbers - you must consider what happens to the units of these numbers. One way to do this is to always try to multiply units. For example, 1 m = 100 cm; I can convert from centimeters to meters or from meters to centimeters by multiplication . To convert from 5 meters to centimeters, I multiply the units (the numbers fall into place): (5 m) X (100 cm / m) = 500 cm (note that the unit m cancelled) To convert from centimeters to meters means inverting the conversion factor: (35 cm) X (1 m / 100 cm) = 0.35 m (the unit cm cancelled) In the problem, you want to convert 0.550 grams B to moles of B. To do this you multiply the units. One is grams and the other is mol / g. (0.550 g B) x (1 mol B / 10.8 g B) = 0.0509 mol B (note the unit g cancels).

For the balanced reaction 4 NH3 + 3 O2 →2 N2 + 6 H2O (Atomic weights: H = 1.0 g/mol; N = 14.0 g/mol; O = 16.0 g/mol), the mass of oxygen needed to react with 75.0 g of ammonia is a) 53.0 g b) 106 g c) 141 g d) 188 g

b) 106 g Why: Once a chemical equation is balanced, there are three steps to follow when you are given the mass of one substance and asked for the mass of a second substance in the reaction. Such is the case here where you are given 75.0 g of ammonia and are asked for the mass of oxygen needed to react with this compound. The three steps are (1) convert grams of what you have to moles (of what you have). (2) convert moles of what you have to moles of what you want. (3) convert moles of what you want to grams of what you want. The choice you picked indicates that you were able to do this. (75.0 g NH3) (1 mol NH3 / 17.0 g NH3) (3 mol O2 / 4 mol NH3) ( 32.0 g O2 / 1 mol O2) = 106 g O2

For the equation w Fe₂S₃ + y O₂→ x Fe₂O₃ + z SO₂ to be properly balanced, the values of w, y, x and z must be a) 1, 9, 1 and 3 b) 2, 9, 2 and 6 c) 2, 7, 2 and 6 d) all coefficients are equal to 1

b) 2, 9, 2 and 6 Why: 2 Fe2S3 + 9 O2 → 2 Fe2O3 + 6 SO2, this is the case. Did we balance the equation the same way? Did you note that there were 3 S atoms on the left of the arrow and only 1 on the right and placed a coefficient of 3 in front of the SO2? Did you then count the oxygen atoms - 2 on the left and 9 on the right? Did you then note that you needed 9 oxygen atoms on the left and, thus, placed a coefficient of 4.5 in front of the O2? Then, to get whole numbers, did you double all the coefficients?

The number of moles of hydrogen gas needed to produce 1.40 mol of ammonia gas according to the reaction N2 + 3 H2 → 2 NH3 is a) 0.700 mol b) 2.10 mol c) 0.930 mol d) 4.20 mol

b) 2.10 mol Why: The coefficients in a balanced equation tell us the number of moles of reactants and products. It also tells us the ratios of moles of these reactants and products to each other. For example, in the reaction of nitrogen and hydrogen to produce ammonia, N2 + 3 H2 → 2 NH3, the equation tells us that 3 moles of hydrogen is required to produce 2 moles of ammonia. From this we can write a conversion factor and use it as follows: (1.40 mol NH3) X (3 mol H2 / 2 mol NH3) = 2.10 mol H2.

The balanced equation for the burning of iron(II) sulfide is: 4 FeS + 7 O2 → 2 Fe2O3 + 4 SO2 The mass of sulfur dioxide produced when FeS (assume you have all that you need) is burned in the presence of 250.0 g of oxygen is (Molar masses: O2 = 32.0 g/mol and SO2 = 64.0 g/mol) a) 500 g b) 286 g c) 143 g d) 875 g

b) 286 g Why: Once a chemical equation is balanced, there are three steps to follow when you are given the mass of one substance and asked for the mass of a second substance in the reaction. Such is the case here where you are given 250.0 g of oxygen and are asked for the mass of sulfur dioxide produced from this compound. The three steps are (1) convert grams of what you have to moles (of what you have). (2) convert moles of what you have to moles of what you want. (3) convert moles of what you want to grams of what you want. The choice you picked indicates that you were able to do this: (250.0 g O2) (1 mol O2/ 32.0 g O2) (4 mol SO2 / 7 mol O2) ( 64.0 g SO2 / 1 mol SO2) = 286 g SO2

How many grams of chromium can be obtained from 75.0 g of sodium dichromate, Na2Cr2O7 (molar mass = 262.0 g/mol)? a) 13.2 g b) 29.8 g c) 39.7 g d) 104 g

b) 29.8 g Why: 75gNa₂Cr₂O₇ X 1 Mol/ 262.0gNa₂Cr₂O₇ = .286 Mol X 104gCr = 29.7gCr

The theoretical yield of Cl2 that could be prepared by mixing 15.0 g of manganese dioxide (MnO2) with 30.0 g of HCl is 12.2 g. The limiting reagent was MnO2. When the reaction was run only 8.82 g of Cl2 was collected. The percent yield of Cl2 is a) 58.8 % b) 72.3 % c) 29.4 % d) 81.3 % e) unknown as there is not enough data to determine the percent yield

b) 72.3 % Why: Percent yield is defined as the actual yield divided by the theoretical yield, then times 100: % yield = (AY/TY) x 100 This problem deals with grams of reactants and products. You realized that the actual yield was 8.82 g of chlorine while the theoretical yield was 12.2 g of chlorine. Thus % yield = (8.82 g/12.2 g) x 100 % yield = 72.3 %

How many moles of sulfur are there in 0.210 mol K3Ag(S2O3)2? a) 4 mol b) 0.0525 mol c) 0.840 mol d) 0.210 mol e) Answer is not a choice

c) 0.840 mol Why: You can use the conversion factor of 4 mol S / 1 mol K3Ag(S2O3)2. Thus, (0.210 mol K3Ag(S2O3)2) x (4 mol S / 1 mol K3Ag(S2O3)2) = 0.840 mol S

For the balanced reaction below: 2 Al + Co2S3 → Al2S3 + 2 Cо the number of moles of Co produced when 100.0 g of Co2S3 reacts with excess Al is (Atomic weight of Co = 58.93 u, Molar mass of Co2S3 = 214.07 g/mol) a) 0.467 mol b) 0.236 mol c) 0.934 mol d) 55.1 mol

c) 0.934 mol Why: The coefficients in a balanced equation tell us the number of moles of reactants and products. It also tells us the ratios of moles of these reactants and products to each other. For example, in the reaction given the equation tells us that 2 moles of cobalt metal are produced per mole of cobalt(III) sulfide reacted. From this we can write the conversion factor 2 mol Co / 1 mol Co2S3. If you start with grams of Co2S3 then you first must convert this to moles Co2S3: (100.0 g Co2S3)x(1 mol Co2S3/214.07 g Co2S3)x(2 mol Co /1 mol Co2S3) = 0.934 mol Co

How many atoms of silver (Ag, atomic weight = 107.9 u) are there in a sample of mass 325 mg? a) 5.00 x 10-26 atoms b) 2.11 x 1025 atoms c) 1.81 x 1021 atoms d) 1.81 x 1024 atoms e) 2.00 x 1023 atoms

c) 1.81 x 1021 atoms Why: To determine the number of silver atoms, you need to calculate the number of moles of silver. For this you will need the mass of silver as well as its molecular mass. These are given in the problem. Next, you will need to use Avogadro's number to get to atoms. First, note that the mass is given in milligrams. You correctly converted this to grams: (325 mg) x (1.00 g / 1000 mg) = 0.325 g Ag (0.325 g Ag) x (1 mol Ag / 107.9 g Ag) = 0.00301 mol Ag (0.00301 mol Ag) x (6.02 x 1023 atoms/mol) = 1.81 x 1021 atoms Ag.

When the equation Mg3N2 + H2O → Mg(OH)2 + NH3 is properly balanced, the sum of the coefficients is a) 11 b) 9 c) 12 d) 29

c) 12 Why: Mg3N2 + 6 H2O → 3 Mg(OH)2 + 2 NH3 is correct. I balanced Mg and N atoms first. I then counted the oxygen atoms and got an answer of 6 (3 x 2 units of the OH ion). I placed a coefficient of 6 in front of the water. Lastly, I counted hydrogen atoms: 12 on the left and 3 x 2 = 6 plus 2 x 3 = 6 for a total of 12 on the right. Lastly, I added the coefficients: 1 + 6 + 3 + 2 = 12!

The formula mass of the ionic compound magnesium phosphate, using the periodic table in the courseware, is a) 119.28 g/mol b) 167.89 g/mol c) 262.87 g/mol d) 214.25 g/mo

c) 262.87 g/mol

Pure chromium metal is prepared from the ore chromite. The ore is converted into sodium dichromate, Na2Cr2O7, which is then reacted with carbon to form the oxide Cr2O3. Reaction of the oxide with aluminum metal yields the chromium metal Cr₂O₃ + 2 Al → Al₂O₃ + 2 Cr If you reacted 50.0 g of Cr2O3 (molar mass = 152.0 g/mol) with 20.0 g of aluminum (molar mass = 27.0 g/mol), the theoretical yield of chromium metal (molar mass = 52.0 g/mol) would be a) 68.4 g b) 17.1 g c) 34.2 g d) 38.5 g

c) 34.2 g Why: In trying to determine the moles of a product in a reaction where you are given masses or moles of different reactants, do you just choose the reactant with the smallest mass or the one with the least number of moles? The answer (and you knew it) is no - you cannot just look for the smallest mass or the least number of moles. For this problem, you need to consider the yield of chromium metal you can get from each reactant. (50.0 g Cr2O3) ( 1 mol Cr2O3 /152 g Cr2O3) (2 mol Cr /1 mol Cr2O3) (52.0 g Cr / mol Cr) = 34.2 g Cr (20.0 g Al) (1 mol Al / 27.0 g Al) (1 mol Cr / 1 mol Al) (52.0 g Cr / mol Cr) = 38.5 g Cr The smallest yield of Cr is 34.2 g. This means that Cr2O3 is the limiting reagent. When all the Cr2O3 is used up we have formed 34.2 g of chromium metal. The other reagent (Al) has not been used up yet. Aluminum is present in the excess (the excess remains unreacted when the reaction is over).

How many grams of zinc (Zn) are there on 0.760 mol of zinc (the atomic mass of zinc is 65.4 u)? a) 86.0 g b) 0.0116 g c) 49.7 g d) 65.4 g

c) 49.7 g Why: The ability to convert grams to moles or the reverse is critical to understanding chemical reactions. Just as we would use the conversion factor 12 inches/foot, or 12 in/ft, to convert a given number of feet (say your height) to inches, a similar conversion factor for grams to moles can be used in chemical examples. The conversion factor is the molar mass of the substance under consideration. For an element the molar mass is simply the atomic mass expressed in units of grams per mole (or g/mol). For zinc, the molar mass is 65.4 g/mol. A second feature in doing conversions is that all units except that wanted must cancel when doing the calculation. You don't just multiply or divide numbers - you must consider what happens to the units of these numbers. One way to do this is to always try to multiply units. For example, 12 in = 1 ft; I can convert from feet to inches or from inches to feet by multiplication . To convert from 5 feet to inches, I multiply the units (the numbers fall into place): (5 feet) X (12 in/ft) = 60 in (note the unit ft cancelled).In the problem, you want to convert 0.760 mol Zn to grams of Zn. To do this you multiply the units. One is moles and the other is g/mol. (0.760 mol Zn) x (65.4 g Zn/mol Zn) = 49.7 g Zn (note the unit mol cancels).

When the following equation Ba(OH)2 + H3PO4 → Ba3(PO4)2 + H2O is correctly balanced, the coefficient in front of the H2O is a) 4 b) 3 c) 6 d) 5 e) 12

c) 6 Why: Ba(OH)2 + H3PO4 → Ba3(PO4)2 + H2O I started by balancing the Ba atoms placing a coefficient of 3 in front of the barium hydroxide? I then balanced the P atoms (or the phosphate units) placing a 2 in front of the phosphoric acid. The result to this point: 3 Ba(OH)2 + 2 H3PO4 → Ba3(PO4)2 + H2O Next, I counted the H atoms in both reactants. I got a total of 12 H atoms. I then balanced these 12 H atoms with 6 water molecules.

Which statement below regarding the concept of a mole in chemistry is not correct? a) One mole of any substance contains the same number of formula units as the number of atoms in exactly 12 g of carbon-12. b) One mole of a substance has a mass in grams that is numerically equal to its formula or molecular mass. c) A mole of water (H2O) contains fewer molecules than a mole of ethanol (C2H5OH). d) All three statements are correct.

c) A mole of water (H2O) contains fewer molecules than a mole of ethanol (C2H5OH). Why: One mole of any substance contains the same number of units (meaning atoms for elements and formulas or molecules for compounds). One mole of water, one mole of ethanol, and one mole of sugar (C12H22O11) all contain the same number of molecules. This number is well known to be 6.022 x 1023 (Avogadro's number).

The empirical formula of an organic compound with the following percent composition: 54.6% C, 9.15% H and 36.2% O is a) C₆HO₄ b) C₅H9O₂ c) C₂H₄O d) Answer you get not a choice

c) C₂H₄O Why: The empirical (or simplest) formula gives the smallest whole number ratio of the atoms in a molecule. To determine the empirical formula you need the percentages of each element in the compound as well as the atomic weights of these elements. I start with the percent of each element. Percent means out of every 100 units - the units can be grams, but I find it easier to use atomic mass units (u). Atomic weights for the elements are in u per atom (u/atom). Thus for C, [54.6 u / 12.0 u/atom] = 4.55 atoms C for H, [9.15 u / 1.01 u/atom] = 9.06 atoms H for O, [36.2 u / 16.0 u/atom] = 2.26 atoms O We therefore have 4.55 atoms C : 9.06 atoms H : 2.26 atoms O . The simplest ratio of atoms is found by dividing all three numbers by the smallest one (2.26). This gives a ratio of 2 atoms C to 4 atoms H to 1 atom O . The correct formula is C2H4O.

One of the sodium oxoacid salts of chlorine (there are 4 salts, each containing Na, Cl, and O) was analyzed. A 17.00 g sample was found to contain 3.68 g Na and 5.67 g Cl. The empirical formula of this salt is a) Na2Cl3O4 b) NaClO2 c) NaClO3 d) Na2Cl2O3

c) NaClO3 Why: Did you find that the sample contained 7.65 g of oxygen (17.00 - 3.68 - 5.67)? This is correct. Did you then convert the mass of each element to moles? This is the correct method. (3.68 g Na) x (1 mol Na atoms / 23.0 g Na) = 0.160 mol Na atoms (5.67 g Cl) x (1 mol Cl atoms / 35.45 g Cl) = 0.160 mol Cl atoms (7.65 g O ) x (1 mol O atoms / 16.0 g Na) = 0.478 mol Na atoms Na0.160Cl0.160O0.478= NaClO3

The reaction between silver nitrate (AgNO3) and barium chloride (BaCl2), both dissolved in aqueous solution, yields a precipitate of silver chloride (AgCl) and a solution containing barium nitrate [Ba(NO3)2]. One way to balance this reaction is 2 AgNO3 + BaCl2 → 2 AgCl + Ba(NO3)2 A second way to balance this reaction is 2 AgNO3 + BaCl2 → Ag2Cl2 + Ba(NO3)2 This second method is... a) correct, but only when excess BaCl2 is used b) correct c) not correct

c) not correct Why: The products of this reaction are silver chloride and barium nitrate. The formula for silver chloride is AgCl. This formula is fixed - it cannot be changed to AgCl2, or Ag2Cl, or Ag2Cl2. These formulas will be for different compounds (if, indeed, these even exist - note: AgCl2 does exist). Once the equation is written, you cannot change the formulas of the reactants or the products. The equation can only be balanced by using coefficients in front of each formula.

Which conversion factor could not be used in solving problems involving Al2(CO3)3? a) 1 mol Al2(CO3)3 / 233.991 g Al2(CO3)3 b) 233.991 g Al2(CO3)3 / 1 mol Al2(CO3)3 c) 2 mol Al / 1 mol Al2(CO3)3 d) 1 mol Al2(CO3)3 / 3 mol O e) All these conversion factors can or cannot be used in solving problems involving Al2(CO3)3

d) 1 mol Al2(CO3)3 / 3 mol O Why: The conversion factor is not correct. If it read 1 mol Al2(CO3)3/ 9 mol O, it would be correct. Note that there are 9 oxygen atoms per formula.

For the reaction 2 C2H5OH + 2 Na2Cr2O7 + 16HCl → 3 CH3COOH + 4 CrCl3 + 4 NaCl + 11 H2O the yield (called a theoretical yield) of CrCl3 we can get from the reaction of 14.0 mol of C2H5OH, 10.0 mol of Na2Cr2O7 and 50.0 mol of HCl would be a) 20.0 mol b) 28.0 mol c) 10.0 mol d) 12.5 mol

d) 12.5 mol Why: In trying to determine the moles of a product in a reaction where you are given masses or moles of different reactants, do you just choose the reactant with the smallest mass or the one with the least number of moles? The answer (and you knew it) is no - you cannot just look for the smallest mass or the least number of moles. For this problem, you need to consider the yield of chromium(III) chloride you can get from each reactant. (14.0 mol C2H5OH) (4 mol CrCl3 / 2 mol C2H5OH) = 28.0 mol CrCl3 (10.0 mol Na2Cr2O7) (4 mol CrCl3 / 2 mol Na2Cr2O7) = 20.0 mol CrCl3 (50.0 mol HCl) (4 mol CrCl3 / 16 mol HCl) = 12.5 mol CrCl3 The smallest yield of CrCl3 is 12.5 moles. This means that HCl is the limiting reagent. When all the HCl is used up we have formed 12.5 moles of CrCl3. The other reagents have not been used up yet. They are present in the excess (they remain unreacted when the reaction is over).

For the reaction 2 SO2 + O2 → 2 SO3, you reacted 5 moles of SO2 with 2 moles of O2. You collected 3 moles of the SO3 product. The percent yield is a) 60 % b) 66 % c) 43 % d) 75 %

d) 75 %

An organic compound of empirical formula C2H5NO3 was found to have a molar mass of approximately 273 g/mol. Using atomic masses of C = 12.0 g/mol, H = 1.0 g/mol, N = 14.0 and O = 16.0 g/mol, the molecular formula for this compound can be determined to be a) C₁₀H₁₂N₃O₆ b) C₂H₅NO₃ c) C₆H₁₅N₃O₃ d) C₆H₁₅N₃O₉

d) C₆H₁₅N₃O₉ Why: The molecular mass of a compound is a whole number multiple of the mass of the empirical formula. The empirical formula gives the simplest whole number ratio of the atoms in a molecule. In this problem the empirical formula is C2H5NO3, which has a mass of 91 u. The formula in this answer has a molar mass of 273 u. This is indeed three times the mass of the empirical formula. In addition, the formula C6H15N3O9 is indeed a whole number multiple of the empirical formula of C2H5NO3 .

The limiting reagent is always the reacting compound with a) the smallest mass b) the least number of moles c) both (a) and (b) are correct d) neither (a) nor (b) need be correct

d) neither (a) nor (b) need be correct Why: To find the limiting reagent requires you to convert masses of reactants to moles and then use the balanced equation to determine which reagent is in excess. The other reagent will be the limiting reagent.

An unknown substance was analyzed and the empirical formula CH2O was reported. Which compounds below could not be the unknown? a) formaldehyde, H-CH = O b) acetic acid, CH₃COOH c) glucose, HO-CH₂(CHOH)₄CH = O d) succinic acid, (CH₂)₂(COOH)₂ e) all of the above are possible

d) succinic acid, (CH₂)₂(COOH)₂ Why: The molecular formula of succinic acid is C₄H₆O₄. Since molecular formula and the empirical formula are related by some whole number, this molecule cannot be the unknown. A compound with the molecular formula C₄H₈O₄ would be correct

Using atomic masses of 69.7 g/mol for gallium and 35.5 g/mol for chlorine, the percent chlorine by mass in the compound GaCl3 can be calculated to be a) 40.5 %. b) 33.32 %. c) 75.01 %. d) 50.9 %. e) 60.44 %.

e) 60.44 %. Why: There are 3 Cl atoms in the molecule (3 X 35.5 = 106.5 g/mol) and 1 Ga atom (1 X 69.7 = 69.7 g/mol). The total mass of 1 mole of gallium chloride is thus 106.5 + 69.7 = 176.2 g. The percent chlorine is (106.5 g/mol / 176.2 g/mol) x 100 = 60.44%.

The actual yield for a chemical reaction is rarely the same as the theoretical yield because a) side reactions often occur giving a different product. b) not enough time was allowed for the reaction to go to completion. c) most reagents used in a reaction are impure. d) all three reasons above apply. e) two of the reasons above.

e) two of the reasons above. Why: 1) side reactions often occur giving a different product. 2) not enough time was allowed for the reaction to go to completion.