Download CHEM 1411 EXAM I (Chapters 1, 2, 3): 25

CHEM 1411 PRACTICE EXAM I (Chapters 1, 2, 3) Zumdahl & Zumdahl Chemistry
Multiple Choices: Select one best answer which is shown in bold.
1. How many significant figures are there in the measurement of 6.124 x 10‐5?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5
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Hint: For 9 ed., see sect. 1.5: p. 15, Interactive Example 1.3.
2. How many significant figures are appropriate to show in the result after carrying out the
operation below?
(223.7 + 0.27) ÷ 4.21 =?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5
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Hint: Comprehensive question. For 9 ed., see sect. 1.5: p.p. 15-18, Interactive Example
1.4(b)(c), end-of-chapter Question 23 and Exercise 1.38 (d). You need to memorize all
the rules concerning the significant figures and the rules related to (1) addition or
subtraction and multiplication and (2) division. Many students do not know the differences
between addition/subtraction and multiplication/division.
See two examples below: Remember that the answers obtained from calculators are
always WRONG when considering the significant figures. So when the question combining
addition/subtraction and multiplication/division together, each follows its own rule.
Thus, 223.7 + 0.27 = 223.97 from calculator, which must be corrected with the least
decimal points of the components, and thus it is 224.0. So the question turns to 224.0 ÷
4.21 = 53.20665083 from calculator, which must be corrected with the lease digit of the
significant figures of the components, and thus it is 53.2.
Since this question only asks how many significant figures in the result, thus from the
components, 224.0 and 4.21, we know one has four significant figures and one has three
significant figures. Thus the number of significant figures of the result should follow the
one with the least significant figures, that is, three significant figures.
3. What is the equivalent temperature for 98.6 Fahrenheit in Kelvin?
(a) 37
(b) 310
(c) 471.6
(d) 594.8
(e) ‐273.1
Hint: For 9th ed., see section 1.8, Interactive Example 1.11. Also see Interactive Example
1.13 as these two examples carry the same concept. There is no direct conversion
between the K and oF. So in this question, you must convert oF to oC first and then convert
o
C to K by applying two formulas: one is oC = (5/9) x (oF – 32) and the other is K = 273.15 +
o
C.
4. Which of the following is a heterogeneous mixture?
(a) benzene and hexane
(b) salt, water, and sugar
(d) 14‐K gold ring
(e) air
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(c) oil and alcohol
Hint: For 9 ed., section 1.10: p. 28. Heterogeneous mixture (having visibly
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distinguishable parts) indicates there are two or more phases in the mixture/solution. If
the mixture is made by solid and liquid, seeing the solid indicates it’s the heterogeneous
mixture. If the mixture is made by liquid and liquid, seeing drops (by shaking it) or layer
indicating it’s the heterogeneous mixture.
5. The mass and volume of an unknown sample is 13.9 g and 17.4 mL, respectively. What is
its density in g/mL?
(a) 0.798
(b) 1.04
(c) 3.16
(d) 4.62
(e) 5.07
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Hint: For 9 ed.: see section 1.9. Density = Mass/Volume
6. Which of the following is a physical change?
(a) water evaporates
(b) wine turns sour (c) scramble an egg
(d) color of carpet faded by sun light
(c) zinc strip dissolves in vinegar to produce hydrogen gas
Hint: For 9th ed., see section 1.10, p. 29. Physical change is the change using the physical
properties, which are the ones you can get them back by changing the temperature or
pressure. The procedure used by adapting physical properties is called the physical process.
Chemical properties are the ones you cannot get them back.
7. How many protons, electrons, and neutrons are there in 72Br ‐ ?
(a) 35p, 35e, 72n
(b) 34p, 35e, 37n
(c) 35p, 36e, 72n
(d) 36p, 34e, 72n
(e) 35p, 36e, 37n
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Hint: Comprehensive question. For 9 ed., see Section 2.5 and 2.6, p.p. 56-57. End-ofChapter Exercise 64, 67, 68, 103.
For negatively charged ion (that is, anion), the negative charge indicates that there is/are
more electrons than protons; while the positively charged ion (that is, cation), the positive
charge indicates there are less electrons than the protons. This is because neutron does
not carry charge, a proton carries one positive charge unit, and an electron carries one
negative charge unit.
The ion in this question carries one negative charge indicating that the electron is one
more than the protons. From periodic table, the atomic number (that is, the proton
number) of Br is 35, and thus there are 35‐(‐1) = 36 electrons.
8. Which of the following is inaccurate?
(a) silicon, Si (b) mercury, Hg
(c) chromium, Cr
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(d) silver, Ag (e) iron, Ir
Hint: For 9 ed., see Section 2.7 or periodic table and memorize them.
9. Which of the following pairs would have similar chemical properties?
(a) Ni, Mg
(b) H, Na
(c) C, Si
(d) Cu, Ca
(e) Cr, Cl
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Hint: For 9 ed., see Section 2.7: p. 59. For 9 edition: p.p. 50‐51. Be careful that H is a
nonmetal while Na is a metal.
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10. Which of the following is a molecular compound?
(a) KCl
(b) CsF
(c) HCN
(d) AlBr3
(e) NaOH
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Hint: For 9 ed., see Section 2.8, p.p. 66‐67; Molecular compound (p. 66) is composed by
nonmetal reacts with nonmetal atoms; while the ionic compound (p. 61) is composed by
metal and nonmetal atoms. However, there is an exception: For compounds that contain
ammonium ion, NH +, they are the ionic compounds as they are formed by acid, HCl, reacts
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with base, NH3.
11. Which of the following is not a correct match?
(a) AlCl3, aluminum chloride
(b) CaH2 calcium hydride
(c) CrF3, chromium (III) fluoride
(d) HNO3, nitrous acid
(e) CuSO4.5H2O, copper (II) sulfate pentahydrate
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Hint: For 9 ed., see Section 2.8: p.p. 60‐70. Interactive Example s 2.3 & 2.5 & End-ofChapter Exercise 73, 79, 83, 85, 88, and 95. You also need to know the name of hydrates
and anhydrous (in the lab manual). Hydrates are the ionic compounds that contain the
crystal water; crystal water is part of the chemical structure/formula; Anhydrous is the
hydrate whose crystal water has been removed. For example, CuSO4 is an anhydrous and is
called “copper (II) sulfate” or “cupric sulfate”.
Follow the rules of “Naming Simple Compounds” Be able to differentiate the rules between
ionic and molecular compounds. Prefix must be included in the name of molecular
compound except the first element with subscript as one. For ionic compound whose
metal is a transition metal or with multiple charges must use Roman numeral to specify the
charge on the metal ion.
H2O is water; Nitric acid is HNO3; Nitrous acid is HNO2
12. Which of the following pairs is a correct match?
(a) Ba3(PO4)2, barium (II) phosphate
(b) (NH4)2SO4, diammonium sulfate
(c) Na2O2, sodium oxide
(d) Ca(NO3)2, calcium nitrate
(e) CH3COOH, carbonic acid
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Hint: For 9 ed., see Section 2.8: p.p. 60‐70. End-of-Chapter Exercise 73, 79, 83, 85, 88 and
95. See “Hint” in Question 11. The correct answer for (a) is barium phosphate because Ba is
a representative metal; for (b) the correction answer is ammonium sulfate because it is an
ionic compound, the exception! The correction answer for (c) is sodium peroxide; while
Na2O is sodium oxide; and for (e) the correction answer is acetic acid or vinegar; Carbonic
acid is H2CO3.
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13. The atomic masses of B and B are 10.0129 u (here “u” stands for “amu”: atomic
mass unit; abundance 19.78%; here “abundance” stands for “natural abundance”) and
11.0093 u (abundance 80.22%), respectively. What is the average atomic mass of B?
(a) 9.467
(b) 9.966
(c) 10.042
(c) 10.504
(e) 10.810
Hint: For 9th ed., see Section 3.2: p.p. 84‐85, Example 3.1. You can convert the % into
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decimals (note: 1% = 0.01) first and then apply the formula: M is the average atomic mass
(or say the atomic mass or atomic weight) which is always shown in the periodic table,
M1 is the mass of isotope # 1 and X1 is the natural abundance (or say the percent
abundance) of isotope #1 in decimal; M2 is the mass of isotope # 2 and X2 is the
natural abundance (or say the percent abundance) of isotope #2 in decimal;
etc. Converting to decimal is to avoid forgetting divided the answer by 100.
For two isotopes: X1 + X2 = 100% =1 always; M = M1X1 + M2X2 = 10.0129x0.1978 +
11.0093x0.8022 = 10.810. Note that the average atomic mass (with decimal) can be found
in the periodic table, while the mass number (a whole number) cannot be found in the
periodic table because it is the sum of the number of protons and the number of neutrons.
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For three isotopes: See End-of-Chapter Exercise 40: 28Si (27.98 u; ? % abundance), Si (?? u;
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4.70% abundance) and Si (29.97 u; 3.09% abundance).
Here ? = 100% - 4.70% - 3.09% = 92. 21%;
The atomic mass M = M1X1 + M2X2 + M3X3 
28.09 = 27.99x0.9221 + ?? x0.0470 + 29.974x0.0309 ?? = 28.98 u
For three isotopes: also see End-of-Chapter Exercise 44.
For five isotopes, see End-of-Chapter Exercise 38. M = M1X1 + M2X2 + M3X3 + M4X4 + M5X5
14. How many moles of S are there in 64.2 g of S?
(a) 1.325
(b) 1.764
(c) 1.9681
(d) 2.003
(e) 2.475
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Hint: For 9 ed., section 3.3: Interactive Example 3.3. Mole = mass (g) / molar mass (g/mole)
= 64.2/32 = 2.003
15. What is the molar mass for calcium nitrate, Ca(NO3)2?
(a) 44
(b) 56
(c) 87
(d) 93
(e) 164
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Hint: For 9 ed., section 3.4: Interactive Exercises 3.6 and 3.7.
In the formula, Ca(NO3)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(NO3)2 = 1x40 + 2x14 + 6x16 = 164
16. How many hydrogen atoms are there in 48.0 g of CH4?
(a) 1.81x1023 (b) 7.22x1024 (c) 6.02x1023 (d) 1.20x10
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(e) 4.70x10
25
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Hint: Hint: For 9 ed., section 3.4: Interactive Exercise 3.8.
According to the chemical formula, one mole of CH4 contains 1 mole of C atoms and 4
moles of hydrogen atoms. Thus, the mole of H = 4 x {mass of CH4/molar mass of CH4}.
When converting moles into atoms, it needs to multiply the Avogadro’s number, that is,
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6.02x10 . Thus the number of H atoms = moles of H atoms x Avogadro’s number =
23
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(6.02x10 )x{4x(48.0/16.0)} = 7.22x10
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17. What is the mass percent (%) for O in SO2?
(a) 38.09
(b) 45.41
(c) 50.00
(d) 53.6
(e) 56.43
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Hint: For 9 ed., section 3.6: Interactive Example 3.9.
% S = (mass of 2 O/ mass of SO2) x 100% = {2x16/(1x32+2x16)}x100% = 50.0%
18. Chemical analysis shows the composition of a compound containing carbon, hydrogen,
chlorine, and oxygen, to be 37.84% C, 2.12% H, 55.84% Cl, and 4.20% O. What is its
empirical formula?
(a) CHClO
(b) C2HClO4
(c) C12H8Cl6O
(d) C12H8Cl6O4
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Hint: For 9 ed., section 3.7: p.p. 96‐103: Interactive Examples 3.10, 3.11, and 3.12.
The empirical formula is the simplest integral ratio of moles among each atom.
Here, there are four different kinds of atoms, C, H, Cl and O.
Thus mole of C = 37.84/12 = 3.15; mole of H = 2.12/1 = 2.12;
mole of Cl = 55.84/35.45 = 1.58; mole of O = 4.20/16 = 0.26.
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 0.26. So C : H : Cl : O = 3.15/0.26 : 2.12/0.26 :
1.58/0.26 : 0.26/0.26 = 12.1: 8.1 : 6.1: 1 = 12: 8: 6: 1 (round to whole number), which
indicates that the empirical formula contains 12 C, 8 H, 6 Cl and 1
O. Thus, the empirical formula is written as C12H8Cl6O as 1 is usually not written in the
formula.
*****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) C4H4O
(b) C2H2O0.5
(c) C8H8O2
(d) CHO
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Hint: For 9 ed., section 3.7: p.p. 96‐103: Interactive Examples 3.10, 3.11, and 3.12. 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 C4H4O as 1 is usually not written in the
formula.
Chemical analysis shows the composition of a compound containing carbon and hydrogen,
to be 80.00% carbon and 20% hydrogen and the molar mass is 30 g. What is its molecular
formula?
(a) CH
(b) C2H4
(c) C2H6
(d) C6H12
(e) C10H22
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Hint: For 9 ed., section 3.7: p.p. 96‐103: Interactive Examples 3.10, 3.11, and 3.12. The
molecular formula is an integral multiple of empirical formula. That is, the molar mass =
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empirical molar mass x integer. From C: H = 80.00/12 : 20.00/1 = 6.66: 20 = 1: 3. So the
empirical formula is CH3 and the empirical molar mass of CH3 = 12x1+1x3 =15. So the
integer = 30/15 = 2. Thus there are two empirical formulas in a molecular formula.
Therefore, the molecular formula is C2H6.
19. Which of the following is the empirical formula, p. 98?
(a) C8H18
(b) CH3COOH
(c) C4H10O2
(d) NH4NO3
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(e) B2H6
Hint: For 9 ed., see section 3.7, p.p. 96-103. p.p. 53‐55. Empirical formula is the simplest
integral ratio among each atom. That is, there should be no common factor among them.
Molecular formula (p. 98) = (empirical formula)n where n is an integer. For (b), the
molecular formula can be written as C2H4O2 and that for (d) as N2H4O3.
20. What are the coefficients respectively when the equation
PH3 + __O2  __ P2O5 + ___ H2O is balanced?
(a) 2, 2, 1, 3
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(b) 2, 1, 3, 4
(c) 2, 3, 1, 2
(d) 2, 4, 1, 3
(e) 1, 1, 3, 3
Hint: For 9 ed., see section 3.9: 105‐108; Interactive Examples 3.13 and 3.14.
Balancing an equation is a process of trial an error. Sometimes it requires more than one
trial. Usually start visual examination and select the most bulky species, that is, the one
with the most different kinds of atoms and number of atoms as the reference and set its
coefficient as one.
In this question, P2O5 is the most bulky one, we put 1 in front of it to remind us we have
done examining P2O5. Now the equation is updated to
PH3 + O2  1 P2O5 + H2O
Since P2O5 contains 2 phosphorus atoms, so we need two phosphorus atoms at the left
side, which leads us to put 2 (called coefficient) in front of the PH3.
Now the equation is updated to 2 PH3 + O2  1 P2O5 + H2O
As there are 6 hydrogen atoms in 2 PH3, thus we need to balance the hydrogen atoms at
the right side, which lead us to put 3 in front of the H2O.
Now the equation is updated to 2 PH3 + O2  1 P2O5 + 3 H2O
Now we need to balance the oxygen atoms. Since there are 1x5+3x1 = 8 oxygen atoms at
the right side, and thus the left side must have the same number of oxygen atoms. That is
to say, x2 = 8. So = 4. So the equation is updated to 2 PH3 + 4 O2 1 P2O5 + 3 H2O
Since all the atoms of each type have the same amount, this equation is balanced.
21. In the reaction of Al(OH)3 with H2SO4, how many moles of water can be produced If the
reaction is begun with 5.500 mole of Al(OH)3?
2Al(OH)3 + 3H2SO4  Al2(SO4)3 + 6H2O
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(a) 2.50
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(b) 4.75
(c) 6.32
(d) 7.58
(e) 16.50
Hint: For 9 ed., see section 3.10: p.p. 108‐114, problem strategy p. 111. It can be
solved by Dimensional Analysis or Unit Analysis.
Stoichiometrey. Method One (ratio approach): From the equation, the involving species,
2 Al(OH)3 and 6 H2O with coefficients 2 and 6 respectively, tell us that for 2 moles of
Al(OH)3 it produces 6 moles of H2O. According to this proportion or ratio, 5.500 mole of
Al(OH)3 produce 5.500 x (6/2) = 16.5 moles of H2O.
Method Two (Problem Solving Strategy, p. 111)
Mass in gram of known substance  moles of known substance moles of desired
substance mass in gram of desired substance .
Grams of known substance ÷ molar mass of substance A = moles of substance A.
Moles of known substance × (coefficient of desired substance ÷ coefficient of known
substance ) = moles of desired substance .
Grams of desired substance = moles of desired substance × molar mass of desired
substance
Since this question starts at moles and thus
22. How many grams of H2O could be formed by the reaction of 16.0 g of CH4 with 48.0 g of
O2?
CH4 + 2O2  CO2 + 2H2O
(a) 27.0
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(b) 37.3
(c) 46.8
(d) 54.1
(e) 58.7
Hint: For 9 ed., see Road Map “Problem-Solving Strategy” in p. 123 and section 3.11: p.p.
114‐121: Interactive Example 3.17.
This question provides two known quantities of substances and thus this is the limiting
reagent question. So we need to apply two times of “Problem Solving Strategy” to figure
out what is the true mass of water. Note that the limiting reagent limits the (maximum) of
product(s) that can be produced. Application, the one produces the smallest amount of the
product is the true limiting reagent. The one with the left over is the excess reagent.
Assume CH4 (i. e. substance A # 1) is the limiting reagent:
÷ 16 g/mol
x 2H2O/1CH4
x 18 g/mol
16 g CH4 ‐‐‐‐‐‐‐‐‐‐‐‐‐>1 mole CH4 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐>2 mole H2O ‐‐‐‐‐‐‐‐‐‐‐‐---‐>36 g H2O
Assume O2 is the limiting reagent:
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÷ 32 g/mol
x 2H2O/2O2
x 18 g/mol
48 g O2 ‐----‐‐‐‐‐‐‐‐‐‐‐-‐>1.5 mole O2 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐---‐> 1.5 moles H2O ‐‐-------‐‐‐‐‐‐‐‐‐>27 g H2O
Because O2 produces the least amount of H2O (can compare by using mole, 1.5 < 2, or
grams, 27 < 36), it is the true limiting reagent and the maximum amount of water
produced is 27 grams. The CH4 is the excess reagent.
To calculate how many grams of the excess reagent were consumed, we apply the road
map:
÷ 32 g/mol
x 1CH4/2O2
x 16 g/mol
48 g O2 ‐‐‐‐‐‐‐‐‐‐‐-----> 1.5 mole O2 ‐‐‐‐‐‐‐‐-------‐‐‐> 0.75 mole CH4‐‐‐‐‐‐‐------‐‐‐‐‐‐> 12 g CH4
Thus, there are 12 g CH4 consumed and there are (16 – 12) = 4 grams of CH4 left.
This is because # grams excess = # grams original – # grams used.
Or say # moles excess = # moles original – # moles used.
23. According to Q 22, which is the limiting reagent? Which is the excess reagent?
(a) Both CH4 and O2 are the limiting reagent.
(b) Both CH4 and O2 are the excess reagent.
(c) CH4 is the excess reagent; O2 is the limiting reagent.
(d) CH4 is the limiting reagent; O2 is the excess reagent.
Hint: See Hint in Q 22.
24. A 15.6 grams of benzene (C6H6) is mixed with excess HNO3 to prepare nitrobenzene
(C6H5NO2). After the reaction there are 13.8 grams of nitrobenzene produced. What is the
percent yield of nitrobenzene?
C6H6 + HNO3 6H5NO2 + H2O
(a) 34.3%
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(b) 47.6%
(c) 56.1%
(d) 63.4%
(e) 71.2%
Hint: For 9 ed., Section 3.11: p.p. 121‐123: Interactive example 3.18 and Road Map
“Problem-Solving Strategy” in p. 123. Here the
The amount of product calculated according to the road map or balanced chemical
equation is the theoretical yield and the one obtained by weighing is the actual yield.
% yield = (actual yield/ theoretical yield) x 100%
Now from the road map, we calculate the theoretical yield (note that since the actual yield
given is in unit of grams, so we need to complete the entire road map).
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÷ 78 g/mol
x 1C6H5NO2 /1C6H6
15.6 gC6H6‐‐‐‐‐‐‐‐‐‐‐‐--------0.2mole C6H6 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐---‐‐‐‐‐‐‐‐------0.2mole C6H5NO2
x 123g/mol
0.2mole C6H5NO2 ‐‐‐‐‐‐‐‐--------------‐-24.6g C6H5NO2
Thus the percent yield = {13.8 g / 24.6 g} x 100% = 56.10%
25. According to Q 24, what is the theoretical yield for nitrobenzene? What is the actual
yield for nitrobenzene?
(a) Theoretical yield: 15.6 g; Actual yield: 15.6 g
(b) Theoretical yield: 24.6 g; Actual yield: 13.8 g
(c) Theoretical yield: 24.6 g; Actual yield: 15.6 g
(d) Theoretical yield: 15.6 g; Actual yield: 24.6 g
(e) Theoretical yield: 13.8 g; Actual yield: 15.6 g
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