CHEMISTRY 181 – UKZN - 2011
TUTORIAL 6
1.
STOICHIOMETRY AND SOLUTION CHEMISTRY
Consider the degradation of glucose to carbon dioxide and water:
C 6 H 12 O 6 + 6O 2 → 6 CO 2 + 6 H 2 O
If 856 g of C 6 H 12 O 6 is consumed by a person over a certain period of
time, what is the mass of CO 2 produced?
Ans: Stoichiometry
nC H
6
1
12O6
nC H
6
nCO
1
2
=
12O6
nCO
6
2
856 g
= 180.2
g/mol
= 4.750 mol C6H12O6
= 4.750 mol x 6 = 28.50 mol CO2
Mass of CO2 = 28.50 mol CO2 x 44.01 g mol-1
= 1.254 x 103 g
2.
The reaction between Li(s) and H2O(ℓ) to produce H2(g) is:
2Li(s) + 2H2O(ℓ) → 2LiOH(aq) + H2(g)
How many grams of Li are needed to produce 9.89 g of H2(g)?
Ans. Conversion steps grams of H2 → moles of H2 → moles of Li → grams of Li
nLi = nH
2
1
2
therfore
nLi = nH2 x 2
=
9.89 g
2.016 g/mol
x2
= 9.812 mols
Mass of Li = 9.812 mols x 6.941 g mol-1 = 68.1 g
3.
Titanium metal is prepared by the reaction:
TiCl4(g) + 2 Mg(ℓ) → Ti(s) + 2MgCl2(ℓ)
In a certain industrial operation that makes aircrafts parts, 3.54 × 107g
of TiCl4 are reacted with 1.13 × 107g of Mg.
(a)
Show using appropriate calculations which is the limiting reagent.
Ans: Calculating the no of mols of product Ti from both reactants
Note: the product MgCl2 can also be used instead of Ti
3.54 x 10 g
nTi = nTiCl /1 = 189.7
g mol
7
4
1.13 x 10 g
nTi = nMg/2 = 24.31
g mol
7
-1
-1
×
×
1
1
= 2.32 × 105 mol
1
= 1.87 x 105 mol
1
therefore TiCl4 is the limiting reagent
(b)
Calculate the theoretical yield of Ti in grams.
Ans: Use the limiting reagent to calculate the theroretical yield
3.54 × 10 g
nTi = nTiCl /1 = 189.7
g mol
4
(c)
7
-1
×
1
1
= 1.87 × 105 mol
mass of Ti = 1.87 × 105 mol × 47.88 g mol-1 = 8.95 × 106 g
Calculate the percentage yield if 7.91 x 106 g of Ti are actually
obtained.
Ans. % Yield =
actual yield
theoretical yield
× 100 =
7.91 x 106 g
8.95 x 106 g
x 100 = 88.4 %
4.
Calculate the mass of solute (in grams) contained in the following
solutions:
(a)
40.0 mL of 0.1265 M KCl
Ans. No. of mols KCl (n) = molarity × volume
= 0.1265 M × 0.040 dm3 (40.0 mL = 0.040 dm3)
= 5.060 × 10-3 mols
Mass of KCl = (5.060 × 10-3 mols ) × 74.55 g mol-1
= 0.3772 g
(b) 12.50 mL of 0.06724 M KMnO4
Ans. No. of mols KMnO4 (n) = molarity × volume
= 0.06724 M × 0.0125 dm3
Mass of KMnO4
5.
= 8.405 × 10-4 mols
= (8.405 × 10-4 mols ) × 158.04 g mol-1
= 0.1328 g
What mass of salt is required to prepare 500 mL of a 100 ppm Cr
solution using K2CrO4.
Ans. 100 ppm = 100 mg K2CrO4 per 1000 mL solvent
% K in K2CrO4 =
molar mass of K x 2
molar mass of K2CrO4
× 100 =
2 x 39.10 g mol-1
194.2 g mol-1
x 100 = 40.28 %
If 100 mg of K2CrO4 is dissolved in 1000 mL solvent , it only constitutes
40.28 % by K.
100 mg = 40.28 % K
x mg = 100 % K
therefore x = 248.7 mg needs to be dissolved in 1000 mL solvent.
Hence 124.35 mg or 1.2435 g of K2CrO4 should be dissolved 500 mL of
solvent to obtain a 100ppm solution.
6.
A solution is prepared by dissolving 0.1164g of (NH4)2Ce(NO3)6 (FW =
548.3g) to 1L with distilled water.
Calculate the ppm of :
(a) NH4+
(a) NH4+ (b) Ce4+ (c) NO3-
-
% NH4+ in (NH4)2Ce(NO3)6 =
18.042 g mol-1
548.3 g mol-1
x 100 = 3.29 %
Mass of NH4+ = 0.0329 × 0.1164g = 3.830 × 10-3 g = 3.83 mg
Therefore conc = 383 mg/L = 383 ppm
(b) Ce4+
% Ce4+ in (NH4)2Ce(NO3)6 =
140.1 g mol-1
548.3 g mol-1
x 100 = 25.55 %
Mass of Ce4+ = 0.2555 × 0.1164g = 7.599 x 10-3 g = 7.599 mg
Therefore conc = 7.599 mg/L = 7.599 ppm
(c) NO3-
% NO3- in (NH4)2Ce(NO3)6 =
6 x 62.01 g mol-1
548.3 g mol-1
x 100 = 67.86 %
Mass of NO3 = 0.06786 × 0.1164g = 7.899 x 10-3 g = 7.899 mg
Therefore conc = 7.899 mg/L = 7.899 ppm
7.
What volume of 16M HNO3 is required to prepare 250 mL of:
(a) 0.01 M HNO3
MHNO3(init) × vol HNO3(init) = MHNO3(final) × vol HNO3(final)
16 M (init) × vol HNO3(init) = 0.01 M (final) × 250 mL (final)
vol HNO3(init) = 0.16 mL
(b) 0.5 M HNO3
MHNO3(init) × vol HNO3(init) = MHNO3(final) × vol HNO3(final)
16 M (init) × vol HNO3(init) = 0.5 M (final) × 250 mL (final)
vol HNO3(init) = 7.81 mL
(c) 6M HNO3
MHNO3(init) × vol HNO3(init) = MHNO3(final) × vol HNO3(final)
16 M (init) × vol HNO3(init) = 6 M (final) × 250 mL (final)
8.
vol HNO3(init) = 93.75 mL
What is the final concentration when 50 mL of a solution of 14 M NH3 is
diluted to:
(a) 100 mL
MNH3(init) × vol NH3(init) = MNH3(final) × vol NH3(final)
14 M (init) × 50 mL(init) = MNH3 (final) × 100 mL (final)
MNH3 (final) = 7.0 M
(b) 250 mL
MNH3(init) × vol NH3(init) = MNH3(final) × vol NH3(final)
14 M (init) × 50 mL(init) = MNH3 (final) × 250 mL (final)
(c) 1L
MNH3 (final) = 2.8 M
MNH3(init) × vol NH3(init) = MNH3(final) × vol NH3(final)
14 M (init) × 50 mL(init) = MNH3 (final) × 1000 mL (final)
MNH3 (final) = 0.7 M
9.
“Muriatic acid” (HCl) is sold for cleaning bricks and roof tiles.
(a)
What is the molarity of this solution if 500mL of the concentrated
solution contains 37 % (w/w) HCl and has a density of 1.18 g cm-3.
Molar mass of HCl = 36.5 g mol-1.
Ans. Assume 100g of the solution
Therefore in 100 g of the solution 37 g of HCl is present
g
nHCl = 36.537g mol
-1
= 1.01 mols
From density (ρ) =
mass(g)
vol(ml)
⇒ vol HCl = mass/ρ
= 37 g /1.18 g mL-1
= 31.36 mL
Therefore MHCl =
(b)
n (mols )
vol(dm3)
=
1.01 mol
0.03136 dm3
= 32.2 mol dm-3 or 32.2 M
How would you prepare a 6 M HCl solution from the concentrated
reagent?
Assume preparation of 100 mL
MHCl (init) × vol HCl(init) = MHCl (final) × vol HCl(final)
32.2 M (init) × vol HCl(init) = 6M × 100 mL
MNH3 (final) = 18.63 mL
Dilute 18.63 mL of the concentrated HCl to 100 mL with distilled
water.
10.
Calculate the volume of 0.01 M NaOH that is required to neutralize:
(a)
25.0 mL of 0.1M HCl
Ans. 1NaOH + 1HCl → NaCl + H2O
nNaOH = nHCl
MNaOH x Vol NaOH = MHCl x Vol HCl
Vol NaOH =
25.0 mL x 0.1 M HCl
0.01 M NaOH
Vol NaOH = 250.0 mL or 0.250 dm3
(b)
25.0 mL of 0.1 M H2SO4
Ans. 2NaOH + 1H2SO4 → Na2SO4 + 2H2O
nNaOH H2SO4
=
1
2
MNaOH x Vol NaOH = MH2SO4 x Vol H2SO4 x 2
Vol NaOH =
Vol NaOH =
M H2SO4 x Vol H2SO4 x 2
M NaOH
0.1 M x 25.0 mL x 2
0.01 M
Vol NaOH = 500 mL or 0.05 dm3
(c)
25.0 mL of 0.1M H3PO4
Ans. 3 NaOH + 1 H3PO4 → Na3PO4 + 3H2O
nNaOH
3
=
nH PO
3
1
4
MNaOH x Vol NaOH = MH3PO4 x Vol H3PO4 x 3
Vol NaOH =
Vol NaOH =
M H3PO4 x Vol H3PO4 x 3
M NaOH
0.1 M x 25.0 mL x 3
0.01 M
Vol NaOH = 625 mL or 0.625 dm3