Section 9.5: Stoichiometry of Solutions Tutorial 1...

Copyright © 2011 Nelson Education Ltd. Chapter 9: Solutions and Their Reactions 9.5-1

Section 9.5: Stoichiometry of Solutions Tutorial 1 Practice, page 447 1. (a) Given:

V

SrCl2

= 150 mL

cSrCl

2

= 0.25 mol / L

cNa

2CO

3

= 0.500 mol/L

Required: volume of 0.500 mol/L sodium carbonate, V

Na2CO

3

Solution: Step 1. Convert all volumes of the solutions to litres.

VSrCl

2

= 150 mL !1 L

1000 mL

VSrCl

2

= 0.150 L

Step 2. Write a dissociation equation listing the calculated amounts and the required value(s). SrCl2(aq) + Na2CO3(aq) → SrCO3(s) + NaCl(aq)

VSrCl

2

= 0.150 L

cSrCl

2

= 0.25 mol / L

cNa

2CO

3

= 0.500 mol/L

VNa

2CO

3

Step 3. Determine the amount of strontium chloride by rearranging the equation.

c =n

V

nSrCl

2

= cSrCl

2

VSrCl

2

=0.25 mol

1L ! 0.15 L

nSrCl

2

= 0.037 50 mol [2 extra digits carried]

Step 4. Determine the amount of sodium carbonate.

nNa

2CO

3

= 0.037 50 molSrCl

2

!

1 molNa

2CO

3

1molSrCl

2

nNa

2CO

3

= 0.037 50 mol

Step 5. Determine the volume of sodium carbonate by rearranging the equation.

VNa

2CO

3

=

nNa

2CO

3

cNa

2CO

3

=0.037 50 mol

0.500mol

L

VNa

2CO

3

= 0.075 L

Statement: The volume of 0.500 mol/L sodium carbonate required is 0.075 L or 75 mL.


(b) Given: n

SrCl2

= 0.037 50 mol (from (a))

Required: mass of precipitate strontium carbonate, m

SrCO3

Solution: Step 1. Write a dissociation equation listing the calculated amounts and the required value(s). SrCl2(aq) + Na2CO3(aq) → SrCO3(s) + NaCl(aq)

n

SrCl2

= 0.037 50 mol

Step 2. Determine the amount of precipitate of strontium carbonate.

nSrCO

3

= 0.037 50 molSrCl

2

!

1 molSrCO

3

1molSrCl

2

nSrCO

3

= 0.037 50 molSrCO

3

Step 3. Calculate the mass of precipitate of strontium carbonate.

mSrCO

3

= 0.037 50 mol !147.63 g

1 mol

mSrCO

3

= 5.5 g

Statement: The mass of strontium carbonate precipitate is 5.5 g. 2. (a) Given:

V

KOH= 70.0 mL

cKOH

= 0.80 mol / L

cFe(NO

3)3

= 0.50 mol/L

VFe(NO

3)3

= 40.0 L

Required: mass of iron(III) hydroxide precipitate, m

Fe(OH)3


VKOH

= 70.0 mL !1 L

1000 mL

VKOH

= 0.070 L

VFe(NO

3)3

= 40.0 mL !1 L

1000 mL

VFe(NO

3)3

= 0.040 L

Step 2. Write the dissociation equation listing the calculated amounts and the required value(s). Fe(NO3)3(aq) + 3 KOH (aq) → Fe(OH)3(s) + 3 KCl(aq)

c

Fe(NO3

)3

= 0.50 mol/L c

KOH= 0.80 mol / L

m

Fe(OH)3

V

Fe(NO3

)3

= 0.040 L V

KOH= 0.070 L


Step 3. Determine the amounts.

c =n

V

nFe(NO

3)3

= cFe(NO

3)3

VFe(NO

3)3

=0.50 mol

1L ! 0.040 L

nFe(NO

3)3

= 0.020 mol

nKOH

= cKOH

VKOH

=0.80 mol

1L ! 0.070 L

nKOH

= 0.056 mol

Step 4. Determine the limiting reagent.

nFe(NO

3)3

= 0.056 molKOH

!

1 molFe(NO

3)3

3molKOH

nFe(NO

3)3

= 0.0187 molFe(NO

3)3

[1 extra digit carried]

Therefore, 0.056 mol potassium hydroxide requires 0.019 mol of iron(III) nitrate. Since 0.020 mol of iron(III) nitrate is present, potassium hydroxide is the limiting reagent. Step 5. Use the amount of limiting reagent to determine the amount of iron(III) hydroxide precipitate.

nFe(OH)

3

= 0.056 molKOH

!

1 molFe(OH)

3

3 molKOH

nFe(OH)

3

= 0.0187 molFe(OH)

3

[1 extra digit carried]

Step 6. Determine the mass of iron(III) hydroxide precipitate.

mFe(OH)

3

= 0.0187 mol !106.88 g

1 mol

mFe(OH)

3

= 2.0 g

Statement: The mass of iron(III) hydroxide precipitate is 2.0 g.


Tutorial 2 Practice, page 448 1. (a) Given:

c

BaCl2

= 0.50 mol/L

Required: amount concentration of the anion, c

Cl–

Solution: Step 1: Write the dissociation reaction. BaCl2(s) → Ba2+(aq) + 2 Cl–(aq) Step 2. Convert the concentration of the compound into concentration of anions.

cCl

–= 0.50

molBaCl

2

L!

2 molCl

–

1molBaCl

2

cCl

–= 1.0 mol/L

Statement: The amount concentration of chloride anion is 1.0 mol/L. (b) Given:

c

KOH= 6.0 mol/L

Required: amount concentration of the anion, c

OH–

Solution: Step 1: Write the dissociation reaction. KOH(s) → K+(aq) + OH–(aq) Step 2. Convert the concentration of the compound into concentration of anions.

cOH

–= 6.0

molKOH

L!

1 molOH

–

1molKOH

cOH

–= 6.0 mol/L

Statement: The amount concentration of hydroxide anion is 6.0 mol/L. (c) Given:

c

Al(ClO3

)3

= 0.10 mol/L

Required: amount concentration of the anion,

cClO

3

–

Solution: Step 1: Write the dissociation reaction. Al(ClO3)3(s) → Al3+(aq) + 3 ClO3

–(aq) Step 2. Convert the concentration of the compound into concentration of cations.

cClO

3

–= 0.10

molAl(ClO

3)3

L!

3 molClO

3

–

1molAl(ClO

3)3

cClO

3

–= 0.30 mol/L

Statement: The amount concentration of chlorate anion is 0.30 mol/L.


2. Given: m

(NH4

)2CO

3

= 14.4 g

V

(NH4

)2CO

3

= 100.0 mL

Required: amount concentration of ammonium ion,

cNH

4

+

Solution: Step 1. Convert the volume of the solution to litres.

V(NH

4)2CO

3

= 100.0 mL !1 L

1000 mL

V(NH

4)2CO

3

= 0.1000 L

Step 2. Determine the amount of solute.

n(NH

4)2CO

3

= 14.4 g !1 mol

96.11 g

n(NH

4)2CO

3


Step 3. Determine the amount concentration.

c(NH

4)2CO

3

=

n(NH

4)2CO

3

V(NH

4)2CO

3

=0.149 83 mol

0.1000 L

c(NH

4)2CO

3

= 1.50 mol/L

Step 4. Write a dissociation equation listing the calculated amounts and the required value(s). (NH4)2CO3(s) → 2 NH4

+(aq) + CO32–(aq)

c

(NH4

)2CO

3

= 1.50 mol/L

Step 5. Convert the concentration of the compound into concentration of cations.

cNH

4

+= 1.50

mol(NH

4)2CO

3

L!

2 molNH

4

+

1mol(NH

4)2CO

3

cNH

4

+= 3.00 mol/L

Statement: The amount concentration of the ammonium ion is 3.00 mol/L.


3. Given: V

Na3PO

4

= 1.75 L

c

Na+= 0.25 mol / L

Required: mass of sodium phosphate, m

Na3PO

4

Solution: Step 1. Write a dissociation equation listing the calculated amounts and the required value(s). Na3PO4(s) → 3 Na+(aq) + PO4

3–(aq)

V

Na3PO

4

= 1.75 L c

Na+= 0.25 mol / L

m

Na3PO

4

Step 2. Convert the concentration of the sodium ion into concentration of sodium phosphate.

cNa

3PO

4

= 0.25

molNa

+

L!

1 molNa

3PO

4

3molNa

+

cNa

3PO

4

= 0.083 mol/L


c =n

V

nNa

3PO

4

= cNa

3PO

4

VNa

3PO

4

=0.083 mol

1L !1.75 L

nNa

3PO

4

= 0.146 mol [1 extra digit carried]

Step 4. Determine the mass of sodium phosphate.

mNa

3PO

4

= 0.146 mol !163.95 g

1 mol

mNa

3PO

4

= 24 g

Statement: The mass of sodium phosphate is 24 g. Section 9.5 Questions, page 449 1. (a) Given:

V

AgNO3

= 32.0 mL

cAgNO

3

= 0.100 mol / L

VNaCl

= 25 mL

Required: amount concentration of sodium ions, c

Na+


VAgNO

3

= 32.0 mL !1 L

1000 mL

VAgNO

3

= 0.032 L


VNaCl

= 25 mL !1 L

1000 mL

VNaCl

= 0.025 L

Step 2. Write a dissociation equation listing the calculated amounts and the required value(s). AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

VAgNO

3

= 0.032 L

cAgNO

3

= 0.100 mol / L

VNaCl

= 0.025 L

cNaCl


c =n

V

nAgNO

3

= cAgNO

3

VAgNO

3

=0.100 mol

1L ! 0.032 L

nAgNO

3

= 0.003 20 mol [1 extra digit carried]

Step 4. Determine the amount of sodium chloride.

nNaCl

= 0.003 20 molAgNO

3

!1 mol

NaCl

1 molAgNO

3

nNaCl

= 0.003 20 molNaCl

Step 5. Determine the amount concentration of sodium chloride.

cNaCl

=n

NaCl

VNaCl

=0.003 20 mol

0.025 L

cNaCl

= 0.13 mol/L

Step 6. Write the dissociation reaction for sodium ions. NaCl(aq) → Na+(aq) + Cl–(aq) Step 7. Convert the concentration of the compound into concentration of sodium cations.

cNa

+= 0.13

molNaCl

L!

1 molNa

+

1molNaCl

cNa

+= 0.13 mol/L

Statement: The amount concentration of sodium ions is 0.13 mol/L.


(b) Given: V

NaCl= 0.032 L

n

NaCl= 0.00320 mol / L (from (a), 1 extra digit carried)

Required: amount concentration of sodium chloride in g/L Solution: Step 1. Convert the amount of solute to mass.

mNaCl

= 0.003 20 mol !58.54 g

1 mol

mNaCl

= 0.187 g [1 extra digit carried]

Step 2. Determine concentration of the solute using the mass.

cNaCl

=m

NaCl

VNaCl

=0.187 g

0.025 L

cNaCl

= 7.5 g/L

Statement: The amount concentration of sodium chloride is 7.5 g/L. 2. (a) NiSO4(aq) + 2 NaOH(aq) → Ni(OH)2(s) + Na2SO4(aq) (b) Given:

c

NiSO4

= 0.45 mol / L

VNiSO

4

= 50.0 mL

VNaOH

= 25.0 mL

cNaOH

= 1.00 mol / L

Required: mass of nickel hydroxide precipitate, m

Ni(OH)2


VNiSO

4

= 50.0 mL !1 L

1000 mL

VNiSO

4

= 0.0500 L

VNaOH

= 25.0 mL !1 L

1000 mL

VNaOH

= 0.0250 L

Step 2. Write a dissociation equation listing the calculated amounts and the required value(s). NiSO4(aq) + 2 NaOH(aq) → Ni(OH)2(s) + Na2SO4(aq)

VNiSO

4

= 0.0500 L

cNiSO

4

= 0.45 mol / L

VNaOH

= 0.0250 L

cNaCl

= 1.00 mol/L


Step 3. Determine the amounts of both substances.

c =n

V

nNiSO

4

= cNiSO

4

VNiSO

4

=0.45 mol

1L ! 0.0500 L

nNiSO

4

= 0.0225 mol [1 extra digit carried]

nNaOH

= cNaOH

VNaOH

=1.00 mol

1L ! 0.0250 L

nNaOH

= 0.0250 mol

Step 4. Determine the amount of sodium hydroxide required to react with 0.0225 mol of nickel(II) sulfate solution.

nNaOH

= 0.0225 molNiSO

4

!2 mol

NaOH

1 molNiSO

4

nNaOH

= 0.0450 mol

Since this amount is greater than the amount present, sodium hydroxide is the limiting reagent and nickel(II) sulfate is the excess reagent. Step 5. Determine the amount of nickel hydroxide precipitate.

nNi OH( )

2

= 0.0250 molNaOH

!

1 molNi OH( )

2

2 molNaOH

nNi OH( )

2

= 0.0125 molNi OH( )

2

Step 6. Determine the mass of nickel hydroxide precipitate.

mNi OH( )

2

= 0.0125 mol !92.71 g

1 mol

mNi OH( )

2

= 1.16 g

Statement: The mass of nickel(II) hydroxide precipitate produced in this reaction is 1.16 g. 3. (a) Given:

m

CaCO3

= 15.2 g

V

Na2CO

3

= 200.0 mL

Required: amount concentration of sodium carbonate solution, c

Na2CO

3


VNa

2CO

3

= 200.0 mL !1 L

1000 mL

VNa

2CO

3

= 0.2000 L


Step 2. Determine the amount of calcium carbonate.

nCaCO

3

= 15.2 g !1 mol

100.09 g

nCaCO

3


Step 3. Write a dissociation equation listing the calculated amounts and the required value(s). Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2 NaCl(aq)

V

Na2CO

3

= 0.200 L m

CaCO3

= 15.2 g

n

CaCO3

= 0.151 86 mol


nNa

2CO

3

= 0.151 86 molCaCO

3

!

1 molNa

2CO

3

1 molCaCO

3

nNa

2CO

3

= 0.151 86 mol

Step 5. Determine the amount concentration.

cNa

2CO

3

=

nNa

2CO

3

VNa

2CO

3

=0.151 86 mol

0.2000 L

cNa

2CO

3

= 0.759 mol/L

Statement: The amount concentration of original sodium carbonate solution is 0.759 mol/L. (b) Given:

m

CaCO3

= 15.2 g

cCaCl

2

= 0.500 mol/L

nCaCO

3

= 0.151 86 mol (from (a), 2 extra digits carried)

Required: volume of calcium chloride solution, V

CaCl

Solution: Step 1. Write a dissociation equation listing the calculated amounts and the required value(s). Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2 NaCl(aq)

c

CaCl2

= 0.500 mol/L m

CaCO3

= 15.2 g

n

CaCO3

= 0.151 86 mol

Step 2. Determine the amount of calcium chloride.

nCaCl

2

= 0.151 86 molCaCO

3

!

1 molCaCl

2

1 molCaCO

3

nCaCl

2

= 0.151 86 mol


Step 3. Rearrange the equation and substitute values.

cCaCl

2

=

nCaCl

2

VCaCl

2

VCaCl

2

=

nCaCl

2

cCaCl

2

=0.151 86 mol

0.500mol

L

VCaCl

2

= 0.304 L

Statement: The volume of calcium chloride solution required to produce 15.2 g of calcium carbonate is 0.304 L or 304 mL. 4. Given:

c

HCl= 12.0 mol/L

m

Fe3O

3

= 224 g

Required: volume of hydrochloric acid, V

HCl

Solution: Step 1. Determine the amount of solute from the mass of solute.

nFe

2O

3

= 224 g !1 mol

159.7 g

nFe

2O

3

= 1.4026 mol [2 extra digits carried]

Step 2. Write a dissociation equation listing the calculated amounts and the required value(s). Fe2O3(s) + 6 HCl(aq) → 2 FeCl3(aq) + 3 H2O(l)

m

Fe2O

3

= 224 g c

HCl= 12.0 mol/L

n

Fe2O

3

= 1.4026 mol

Step 3. Determine the amount of hydrochloric acid.

nHCl

= 1.4026 molFe

2O

3

!6 mol

HCl

1molFe

2O

3

nHCl

= 8.4158 molHCl

[2 extra digits carried]


Step 4. Rearrange the equation and substitute values to calculate the volume of hydrochloric acid.

cHCl

=n

HCl

VHCl

VHCl

=n

HCl

cHCl

=8.4158 mol

12.0 mol

L

VHCl

= 0.701 L

Statement: The volume of hydrochloric acid is 0.701 L. 5. (a) 2 Al(s) + 3 CuSO4(aq) → 3 Cu(s) + Al2(SO4)3(aq) (b) Given:

c

CuSO4

= 0.100 mol/L

V

CuSO4

= 150 mL

Required: mass of aluminum, m

Al


VCuSO

4

= 150 mL !1 L

1000 mL

VCuSO

4

= 0.15 L

Step 2. Determine the amount of copper(II) sulfate.

cCuSO

4

=

nCuSO

4

VCuSO

4

nCuSO

4

=VCuSO

4

! cCuSO

4

= 0.15 L !0.100 mol

1 L

nCuSO

4

= 0.015 mol

Step 3. Determine the amount of aluminum.

nAl= 0.015 mol

CuSO4

!2 mol

Al

3 molCuSO

4

nAl= 0.010 mol

Al


Step 4. Determine the mass of aluminum.

mAl= 0.010 mol !

26.98 g

1 mol

mAl= 0.27 g

Statement: The mass of aluminum required to react with the copper(II) sulfate solution is 0.27 g. 6. Given:

m

NaOH= 45!10

6 t

c

NaCl= 6.0 mol/L

Required: volume of sodium chloride, V

NaCl

Solution: Step 1. Convert the mass of sodium hydroxide to grams.

mNaOH

= 45!106

t !10

6 g

1 t

mNaOH

= 45!1012

g

Step 2. Determine the amount of sodium hydroxide.

nNaOH

= 45!1012

g !1 mol

40.00 g

nNaOH

= 1.125!1012

mol [2 extra digits carried]

Step 3. Write a dissociation equation listing the calculated amounts and the required value(s). 2 NaCl(aq) + 2 H2O(l) → 2 NaOH(aq) + Cl2(g) + H2(g)

c

NaCl= 6.0 mol/L

n

NaOH= 1.125!10

12mol

Step 4. Determine the amount of sodium chloride.

nNaCl

= 1.125 !1012

molNaOH

!1 mol

NaCl

1 molNaOH

nNaCl

= 1.125 !1012

molNaCl

Step 5. Rearrange the equation and substitute values to calculate the volume of hydrochloric acid.

cNaCl

=n

NaCl

VNaCl

VNaCl

=n

NaCl

cNaCl

=1.125!10

12mol

6.0 mol

L

VNaCl

= 1.9 !1011

L

Statement: The volume of sodium chloride solution required is 1.9 × 1011 L.


7. Given: c

AgNO3

= 0.50 mol / L

VAgNO

3

= 15.0 mL

msolute

= 0.42 g

Vsolution

= 100.0 mL

Required: identify the solute Solution: Step 1. Convert the volumes of the solutions to litres.

VAgNO

3

= 15.0 mL !1 L

1000 mL

VAgNO

3

= 0.0150 L

Vsolution

= 100.0 mL !1 L

1000 mL

Vsolution

= 0.1000 L

Step 2. Determine the amount of silver nitrate used.

nAgNO

3

= 0.0150 L !0.50 mol

1 L

nAgNO

3

= 7.5!10"3

mol

Step 3. Using sodium chloride as the solute, determine the amount of sodium chloride.

nNaCl

= 0.42 g !1 mol

58.44 g

nNaCl

= 7.2 !10"3

mol

Step 4. Using sodium chloride as the solute, write the dissociation equation. NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3 (aq)

n

NaCl= 7.2 !10

"3mol

n

AgNO3

Step 5. Using sodium chloride as the solute, determine the amount of silver nitrate required.

nAgNO

3

= 7.2 !10"3

molNaCl

!

1 molAgNO

3

1 molNaCl

nAgNO

3

= 7.2 !10"3

mol

Step 6. Using calcium chloride as the solute, determine the amount of calcium chloride.

nCaCl

2

= 0.42 g !1 mol

110.98 g

nCaCl

2

= 3.78 !10"3

mol [1 extra digit carried]

Step 7. Using calcium chloride as the solute, write the dissociation equation. CaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ca(NO3)2(aq)

n

CaCl2

= 3.78 !10"3

mol n

AgNO3


Step 8. Using calcium chloride as the solute, determine the amount of silver nitrate required.

nAgNO

3

= 3.78 !10"3

molCaCl

2

!

2 molAgNO

3

1 molCaCl

2

nAgNO

3

= 7.56 !10"3

mol

Statement: The actual amount of silver nitrate available is 7.5 ×10-3 mol. If the unknown solution contains sodium chloride, it would take 7.2 × 10-3 mol of silver nitrate to precipitate all of the chloride ions. If the unknown solution contains calcium chloride, it would take 7.56 × 10-3 mol of silver nitrate to precipitate all of the chloride ions. The unknown solution is sodium chloride because there is not enough silver nitrate to precipitate all of the chloride ions if the solution was calcium chloride. 8. (a) Given:

c

Na2CO

3

= 0.5 mol / L

Step 1: Write the dissociation reaction. Na2CO3(aq) → 2 Na+(aq) + CO3

2-(aq) Step 2. Convert the concentration of the compound into concentration of cations.

cNa

+= 0.5

molNa

2CO

3

L!

2 molNa

+

1molNa

2CO

3

cNa

+= 1 mol/L

Statement: The amount concentration of cations is 1 mol/L. (b) Given:

c

(NH4

)2

SO3

= 0.2 mol / L

Step 1: Write the dissociation reaction. (NH4)2SO3(s) → 2 NH4

+ + SO32–


cNH

4

+= 0.2

mol(NH

4)2

SO3

L!

2 molNH

4

+

1mol(NH

4)2

SO3

cNH

4

+= 0.4 mol/L

Statement: The amount concentration of cations is 0.4 mol/L. (c) Given:

c

Fe2

(SO4

)3

= 1.5 mol / L

Step 1: Write the dissociation reaction. Fe2(SO4)3(s) → 2 Fe3+ + 3 SO4

2–


cFe

3+= 1.5

molFe

2(SO

4)3

L!

2 molFe

3+

1molFe

2(SO

4)3

cFe

3+= 3.0 mol/L

Statement: The amount concentration of cations is 3.0 mol/L.


9. Given: c

Na+= 0.85 mol/L

V

Na2CO

3

= 200.0 mL

Required: mass of sodium carbonate, m

Na2CO

3


VNa

2CO

3

= 200.0 mL !1 L

1000 mL

VNa

2CO

3

= 0.2000 L

Step 2. Write the dissociation reaction. Na2CO3(aq) → 2 Na+(aq) + CO3

2-(aq) Step 3. Determine the concentration of sodium carbonate.

cNa

2CO

3

= 0.85 mol

Na+

L!

1 molNa

2CO

3

2 molNa

+

cNa

2CO

3

= 0.425 mol/L [1 extra digit carried]


cNa

2CO

3

=

nNa

2CO

3

VNa

2CO

3

nNa

2CO

3

= cNa

2CO

3

!VNa

2CO

3

= 0.425 mol

L! 0.2000 L

nNa

2CO

3

= 0.085 mol

Step 5. Determine the mass of sodium carbonate.

mNa

2CO

3

= 0.085 mol !105.99 g

1 mol

mNa

2CO

3

= 9.0 g

Statement: The mass of sodium carbonate is 9.0 g.

Section 9.5: Stoichiometry of Solutions Tutorial 1...

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