Chapter 3 Answers to Review Problems PDF

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Solutions Manual to accompany

Chemistry 2nd Edition by Blackman, Bottle, Schmid, Mocerino & Wille

Prepared by John Hill

© John Wiley & Sons Australia, Ltd 2012

Solutions Manual to accompany Chemistry 2e by Blackman et al.

Chapter 3 - Chemical reactions and stoichiometry Practice questions 3.1

1 Mg, 2 O, 4 H and 2 Cl (on each side).

3.2

Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)

3.3

3 BaCl2(aq) + Al2(SO4)3(aq)  3BaSO4(s) + 2AlCl3(aq)

3.4

mol S = 

3.5

0.125 mol  106 g mol–1 = 13.3 g

3.6

mol H2SO4 = 

3.7

g Fe = 

3.8

g Fe = 

3.9

%N = 0.1417/0.5462  100 = 25.94% %O = 0.4045/0.5462  100 = 74.06% Since the sum of %N and %O equals 100%, there are no other elements present.

3.10

We first determine the number of grams of each element that are present in one mole of sample: 2 mol N  14.01 g mol–1 = 28.02 g N 4 mol O  16.00 g mol–1 = 64.00 g O



 35.6 g S  = 1.11 mol 1  32.07 g mol S 



45.8 g H 2SO 4   = 0.467 mol 1  98.1 g mol H2SO4 



 2 mol Fe  15.0 g Fe2 O3 1    55.8 g mol Fe = 10.49 g 1 159.7 g mol Fe O 1 mol Fe O 2 3  2 3  

25.6 g O   2 mol Fe  55.8 g mol1 Fe  = 59.5 g    1  16.0 g mol O   3 mol O 

© John Wiley & Sons Australia, Ltd 2012 3.2

Chapter 3: Chemical reactions and stoichiometry The percentages by mass are then obtained using the molar mass of the compound (92.02 g mol–1): %N = 28.02/92.02 = 30.45% %O = 64.00/92.02 = 69.55% 3.11

It is convenient to assume that we have 100 g of the sample, so that the % mass represents masses. Thus there is 32.4 g of Na, 22.6 g of S and (100.00 – 32.4 – 22.6) = 45.00 g of O. Convert these masses to moles:

  32.4 g Na mol Na =   = 1.40 mol 1  23.00 g mol Na    22.6 g S mol S =   = 0.70 mol 1  32.06 g mol S    45.0 g O mol O =   = 2.81 mol 1 16.00 g mol O  Next, we divide each of these mole amounts by the smallest in order to deduce the simplest whole number ratio: For Na: 1.40 mol/0.70 mol = 2.0 For S: 0.70 mol/0.70 mol = 1.0 For O: 2.81 mol/0.70 mol = 4.0 The empirical formula is therefore Na2SO4.

3.12

 1 mol H 2SO 4  mol H 2SO 4 = 0.366 mol NaOH    = 0.183 mol  2 mol NaOH 

3.13

 5 mol O 2  mol O2 =  0.575 mol CO 2    = 0.958 mol  3 mol CO2 

3.14

 1 mol Al2 O3  g Al 2O 3 = 1.54 mol Fe  102.0 g mol 1 Al 2O 3  = 78.5 g    2 mol Fe 

3.15

First determine the mass of O2 that is required to react completely with the given mass of ammonia:

 30.00 g NH 3  5 mol O2  1 g O2 =    32.00 g mol O 2  1 17.03 g mol NH 4 mol NH 3  3  = 70.46 g

© John Wiley & Sons Australia, Ltd 2012 3.3

Solutions Manual to accompany Chemistry 2e by Blackman et al.

Since this is more than the amount that is available, we conclude that oxygen is the limiting reactant, and therefore:

 40.00 g O 2  4 mol NO  1 g NO =     30.01 g mol NO  1  32.00 g mol O2  5 mol O2  = 30.01 g 3.16

First determine the mass of C2H5OH that is required to react completely with the given amount of sodium dichromate:

  3 mol C2 H5OH  90.0 g Na 2Cr2O7 1 g C 2H 5OH =    46.08 g mol C 2H 5OH  1  262.0 g mol Na 2 Cr2 O7  2 mol Na2 Cr2 O7  = 23.7 g Once this amount of C2H5OH is reacted the reaction ceases, even though 24.0 g C2 H5OH remain, because all the Na2Cr2O7 has reacted. Therefore Na2Cr2O7 is the limiting reactant, and the theoretical yield of acetic acid (CH3COOH) is therefore based on the amount of Na2Cr2O7 added:

  3 mol CH3 COOH  90.0 g Na 2Cr2O7 1 g CH3COOH =    60.06 g mol CH3COOH  1 262.0 g mol Na Cr O 2 mol Na Cr O 2 2 7  2 2 7   = 30.9 g Now the percentage yield is calculated from the mass of acetic acid produced, 26.6 g:

 actual yield  percent yield =    100   theoretical yield 

3.17

 26.6 g CH 3COOH    100  86.1%  30.9 g CH3COOH 

 3.550 g Na 2SO 4  mol Na 2SO 4 =   = 0.024 98 mol 1  142.1 g mol Na 2SO4   1L  L solution = 100.0 mL    = 0.1000 L  1000 mL   moles solute   0.02498 mol Na 2SO 4  M=     0.2498 M  L solution   0.1000 L solution 

© John Wiley & Sons Australia, Ltd 2012 3.4

Chapter 3: Chemical reactions and stoichiometry

3.18

3.19

 1 L solution  0.0125 mol AgNO3  g AgNO 3 =  250 mL solution     1 L solution  1000 mL solution   1 169.90 g mol AgNO3  = 0.53 g (Vdil)(Mdil) = (Vconc)(Mconc) (100 mL)(0.125 M) = (Vconc)(0.500 M) Vconc = (100 mL)(0.125 M)/(0.500 M) = 25.0 mL Therefore, mix 25.0 mL of 0.500 M H2SO4 with water to make 100 mL of total solution.

3.20

 1L  1 mol H 2SO 4 = 15.4 mL solution    0.108 mol L = 0.00166 mol 1000 mL 

 2 mol NaOH  mol NaOH =  0.00166 mol H 2SO 4    = 0.00333 mol  1 mol H2 SO4   0.00333 mol NaOH  1000 mL  mL NaOH=    = 26.8 mL 1  0.124 mol L NaOH  1 L  3.21

FeCl3  Fe3+ + 3Cl–

 0.40 mol FeCl3  1 mol Fe 3+   = 0.40 M Fe3+  M Fe3+ =    1 L FeCl3 soln  1 mol FeCl3   0.40 mol FeCl3   3 mol Cl     = 1.2 M Cl    M Cl =    1 L FeCl3 soln  1 mol FeCl3 

3.22

 0.250 mol PO4 3   3 mol Na +  M Na =   1 mol PO 3 1 L Na PO soln 3 4 4  

3.23

The balanced net ionic equation is: Fe2+(aq) + 2OH–(aq)  Fe(OH)2(s). First determine the number of moles of Fe2+ present:

+

  = 0.750 M Na  

0.250 mol FeCl2  1 mol Fe 2+   = 1.50 10 2 mol Fe 2+   1000 mL solution  1 mol FeCl2 

60.0 mL FeCl2 solution

Now, determine the amount of KOH needed to react with the Fe2+:

© John Wiley & Sons Australia, Ltd 2012 3.5

Solutions Manual to accompany Chemistry 2e by Blackman et al.

1.50 10 3.24

2

 2 mol OH mol Fe 2+  2+  1 mol Fe



  1 mol KOH  1000 mL solution     = 60.0 mL KOH     1 mol OH  0.500 mol KOH 

2– The net ionic equation is Ba 2+(aq) + SO4 (aq)  BaSO4(s). First, determine the initial number of moles of Ba 2+ ion that are present:

 0.600 mol BaCl2 (20.0 mL BaCl2 soln) 1000 mL BaCl 2 soln

 1 mol Ba 2+   1 mol BaCl 2

  = 1.20  10 2 mol Ba 2+ 

Next, determine the initial number of moles of sulfate ion that are present: 2  0.500 mol MgSO 4   1 mol SO 4     = 1.50  10 2 mol SO 4 2  (30.0 mL MgSO 4 soln)   1000 mL MgSO 4 soln   1 mol MgSO 4 

Now determine the number of moles of barium ion that are required to react with this amount of sulfate ion, and compare the result to the amount of barium ion that is available: 2+   1 mol Ba (1.50  102 mol SO 42 )  1 mol SO 2  4

  = 1.50  10 2 mol Ba 2 +  

Since there is not this amount of Ba2+ available according to the above calculation, then we can conclude that Ba 2+ must be the limiting reactant, and that subsequent calculations should be based on the number of moles of it that are present: Since this reaction is 1:1, we know that 1.20 × 10–2 mole of BaSO4 will be formed. If we assume that the BaSO4 is completely insoluble, then the concentration of barium ion is zero. The concentrations of the other ions are determined as follows:

 0.600 mol BaCl 2  2 mol Cl    1000 mL BaCl 2 soln  1 mol BaCl 2  (20.0 + 30.0) mL soln) 1 L soln   1000 mL soln 

 20.0 mL BaCl2 soln

   = 0 .480 M Cl 

 0.500 mol MgSO 4   1 mol Mg2      1000 mL MgSO 4 soln   1 mol MgSO 4 

30.0 mL MgSO 4 soln 

 (30.0 + 20.0) mL soln) 1 L soln   1000 mL soln  For sulfate, we subtract the amount that reacted with the Ba 2+: © John Wiley & Sons Australia, Ltd 2012 3.6

= 0.300 M Mg2

Chapter 3: Chemical reactions and stoichiometry 1.50 × 10–2 mol – 1.20 × 10–2 mol = 3.0 × 10–3 mol This allows a calculation of the final sulfate concentration: 2

3.0 10 3 mol SO 4 2 = 6.0  10 2 M SO 4 1 L soln  (30.0 + 20.0) mL soln)  1000 mL soln  

3.25

 1 mol CaSO 4  1 mol Ca 2+   = 5.41 103 mol Ca 2+   136.14 g CaSO 4  1 mol CaSO4 

(a)

0.736 g CaSO 4 

(b)

Since all of the Ca 2+ is precipitated as CaSO4, there were originally 5.41 × 10–3 moles of Ca 2+ in the sample.

Review questions 3.1

A chemical equation is balanced when there is the same number of each kind of atom on both the reactant and product side of the equation. This condition arises from the law of conservation of matter.

3.2

‘Reactants’ are the initial components of a chemical reaction and their chemical formulae are given on the left side of the arrow. ‘Products’ are the final components of the reaction and their chemical formulae are given on the right side of the arrow.

3.3

Coefficients

3.4

(a) (b) (c) (d)

Magnesium reacts with oxygen to give (yield) magnesium oxide. The reactants are Mg and O2. The product is MgO. 2Mg(s) + O2(g)  2MgO(s)

3.5

(a) (b)

Student B is correct. Student A wrote a properly balanced equation. However, by changing the subscript for the product of the reaction from an implied one, NaCl, to NaCl2, this student has changed the identity of the product. When balancing chemical equations, never change the value of the subscripts in chemical formulae.

3.6

The ‘mole’ is the SI unit for the ‘amount’ of a substance. One mole is equal to Avogadro’s constant (6.022  1023) of particles, or the molar mass in grams of a substance.

© John Wiley & Sons Australia, Ltd 2012 3.7

Solutions Manual to accompany Chemistry 2e by Blackman et al.

3.7

‘Mole’ not ‘mass’ is the fundamental unit of chemistry. Stoichiometry is given by a balanced chemical equation, which is directly interpreted in ‘moles’.

3.8

Molar mass

3.9

Amount of substance (moles) 

3.10

To estimate the number of atoms in 1 gram of iron, first convert g to kg and then use the relationship, 1.661  10–27 kg = 1 u, and finally divide by the atomic mass of Fe (55.85 u):  1 kg   1000 g  1.661

1 g Fe 

Mass (g) Molar mass (g mol1 )

  1 molecule  1u 22   = 1.08  10 atoms Fe 27  10 kg   55.85 u 

3.11

There are the same number of molecules in 2.5 moles of H2O and 2.5 moles of H2.

3.12

There are 2 moles of iron atoms in 1 mole of Fe2O3. The number of iron atoms in 1 mole of Fe2O3 is:  2 mol Fe  23 –1 24 (6.022 × 10 mol ) = 1.204  10 atoms Fe 1 mole Fe2O3 

1 mol Fe2O3 

3.13

The statement ‘1.0 mole of oxygen’ does not indicate whether this is atomic oxygen, O, or molecular oxygen, O2. The statement ‘64 g of oxygen’ is not ambiguous because the source of oxygen is not important.

3.14

As a minimum, the identity and mass of each atomic element present must be known. If the total mass of the compound is known, then it is necessary to know all but one mass of the elements that compose the compound.

3.15

When balancing a chemical equation, changing the subscripts of a chemical formula changes the identity of the corresponding compound.

3.16

Convert moles of B to moles of compound A5B2. Then using the stoichiometric ratio of moles of A to moles of A5B2, determine the moles of A and finally convert the moles of A to grams of A using the molar mass of A. 1 mol A B   5 mol A  100.0 g A 5 2       2 mol B  1 mol A5 B2   1 mol A 

10 mol B 

© John Wiley & Sons Australia, Ltd 2012 3.8

Chapter 3: Chemical reactions and stoichiometry Both (iii) and (iv) are unnecessary. Molar mass of B is not needed as the question states the number of moles of B and 6.022 × 1023 atoms constitute1 mole of any substance.

3.17

To determine the number of grams of sulfur that reacts with 1 gram of arsenic, the stoichiometric ratio of arsenic to sulfur in the compound is needed together with the molar masses of sulfur and arsenic.

3.18

(a) (b)

3.19

2 H 2O 2  2 H 2 O + O 2

3.20

Concentration (mol L1 ) 

3.21

Molarity is the number of moles of solute per litre of solution, also known as molar concentration.

The balanced equation describes the stoichiometry of a chemical reaction. The scale of the reaction is determined by the mass of the reactants.

Amount of substance (moles) Volume (L)

 mmol  1 mol  1000 mL   1000 mol   mol   mL  1000 mmol  1 L    1000 L    L          3.22

A molecular equation is a chemical equation that gives the molecular formulas for all reactants and products of a reaction. An ionic equation is a chemical equation in which all of the soluble strong electrolytes are represented in their dissociated form. A net ionic equation shows only those ions and molecules that are chemically involved in the reaction. A net ionic equation differs from an ionic equation in that all of the spectator ions are omitted from the former. Spectator ions do not take part in a reaction. These are ions that result from strong electrolytes.

3.23

On dilution, the number of moles of HNO3 in the solution has not changed but the concentration has decreased since water has been added.

3.24

(VA)(MA) = (VB)(MB) (50 mL)(0.10 M) = (VB)(0.20 M) VB = (50 mL)(0.1 M)/(0.20 M) = 25.0 mL is the volume of solution B

3.25

Qualitative analysis is the use of experimental procedures to determine the elements that are present in a substance.

© John Wiley & Sons Australia, Ltd 2012 3.9

Solutions Manual to accompany Chemistry 2e by Blackman et al.

Quantitative analysis determines the percentage composition of a compound or the percentage of a component in a mixture. Qualitative analysis answers the question, ‘what is in the sample?’ Quantitative analysis answers the question, ‘how much is in the sample?’

3.26

The spectator ions are Na + and Cl–. The net ionic equation is: Co2+ + 2OH–  Co(OH)2(s)

3.27

The charge on Co is incorrect and the physical states of the reactants and products are not 2– + given. Balanced equation is: 3Co2+ (aq) + 2HPO4 (aq)  Co3(PO4)2(s) + 2H (aq).

3.28

(a) (b) (c)

3.29

2(CH3)2S + 902 → 2S02 + 4CO2 + 6H2O

3.30

4Fe(s) + 302(g) → 2Fe2O3(s)

3.31

CH3OH + 3N2O → CO2 + 2H2O + 3N2

3.32

3NO2 + H2O → 2HNO3 + NO

3.33

4 H2NCHO + 5O2 → 4CO2 + 6H2O + 2N2

3.34

4 H2NCHO + 5O2 → 4CO2 + 6H2O + 2N2 (One extra H2O shown in 'Products')

3.35

(a) (b) (c) (d) (e)

3.36

2 molecules of SO2 are formed.

3.37

O2 is the limiting reagent.

3.38

Natom:Oatom is 1:2, therefore Nmol:Omol is 1:2.

6 3 27

Ca(OH)2 + 2HCl  CaCl2 + 2H2O 2AgNO3 + CaCl2  Ca(NO3)2 + 2AgCl 2Fe2O3 + 3C  4Fe + 3CO2 2NaHCO3 + H2SO4  Na2SO4 + 2H2O + 2CO2 2C4H10 + 13O2  8CO2 + 10H2O

© John Wiley & Sons Australia, Ltd 2012 3.10

Chapter 3: Chemical reactions and stoichiometry



 –3  = 2.59  10 mole Na atoms Na 

1 mol Na

3.39

1.56  1021 atoms Na 

3.40

 2 mol Al  mol Al = (1.58 mol O)   = 1.05 mol  3 mol O 

3.41

Based on the balanced equation: 2NH3(g)  N2(g) + 3H2(g)

6.022  1023

Hence, the conversion factors are:

 3 mol H 2   1 mol N 2    and    2 mol NH 3   2 mol NH3  To determine the moles produced, convert ‘starting moles’ to ‘end moles’:

 1 mol N2    0.0725 mol N 2 0.145 mol NH3   2 mol NH3  The moles of hydrogen are:

 3 mol H 2 0.145 mol NH 3   2 mol NH3

3.42

3.43

   0.218 mol H 2 

 4 mol UF6 mol UF6 = 1.25 mol CF4   6 mol CF4

  = 0.833 mol 

The mass of oxygen in the compound is determined by difference: 0.896 g total – (0.111 g Na + 0.477 g Tc) = 0.308 g O. Next we convert each mass into the corresponding number of moles:

 0.111 g Na  moles Na =  = 4.83  10  3 moles  1  23.00 g mol Na   0.477 g Tc  3 moles Tc =   = 4.82  10 moles 1  98.9 g mol Tc   0.308 g O  2 moles O =   = 1.93  10 moles 1  16.0 g mol O  Now we divide each of these numbers of moles by the smallest of the three, in order to obtain the simplest mole ratio of the three elements in the compound: for Na, 4.83  10–3 moles / 4.82  10–3 moles = 1.00 © John Wiley & Sons Australia, Ltd 2012 3.11

Solutions Manual to accompany Chemistry 2e by Blackman et al.

for Tc, 4.82  10–3 moles / 4.82  10–3 moles = 1.00 for O, 1.93  10–2 moles / 4.82  10–3 moles = 4.00 Thus the empirical formula is NaTcO4. 3.44

Assume a 100g sample:

  14.5 g C mol C     1.21 mol 1  12.01 g mol C    85.5g Cl mol Cl     2.41 mol 1 35.45 g mol Cl   Now we divide each of these numbers of moles by the smallest of the two, in order to obtain the simplest mole ratio of the two elements in the compound: for C, 1.21 moles / 1.21 moles = 1.00 for Cl, 2.41 moles / 1.21moles = 2.000 Thus the empirical formula is CCl2. 3.45

All of the carbon is converted to carbon dioxide so:

 1.312 g CO 2  1 mol C  mol C      0.0298 mol 1  44.01 g mol CO2  1 mol CO2   1.312 g CO 2  1 mol C  1 gC     (12.01 g mol C)  0.358 g 1  44.01 g mol CO2  1 mol CO2  All of the hydrogen is converted to H2O so:

 0.805 g H 2O  2 mol H  mol H      0.0893 mol 1  18.02 g mol H 2O  1 mol H2O   0.805 g H 2O  2 mol H  1 gH     (1.008 g mol H)  0.901 g 1  18.02 g mol H2O  1 mol H2 O  The amount of O in the compound is determined by subtracting the mass of C and the mass of H from the sample mass: g O = 0.684g – 0.358 g – 0.0901 g = 0.236 g

 1 mol O  2 mol O  (0.236 g O)    1.48  10 mol 16 g O   The relative mole ratios are: for C, 0.0298 moles / 0.0148 moles = 2.01 © John Wiley & Sons Australia, Ltd 2012 3.12

Chapter 3: Chemical reactions and stoichiometry for H, 0.0893 moles/ 0.0148 moles = 6.03 for O, 0.0148 moles / 0.0148 moles = 1.00 Thus the empirical formula is C2H6O. 3.46

From the information provided, the mass of mercury is the difference between the total mass and the mass of bromine:

g Hg = 0.389 g compound  0.111 g Br = 0.278 g To determine the empirical formula, first convert mass to moles.

  0.278 g Hg 3 moles Hg =   = 1.39  10 moles 1  200.59 g mol Hg    0.111 g Br 3 moles Br =   = 1.39  10 moles 1 79.904 g mol Br   Now, we divide e...