Chemistry 1000 (Spring 2006)
Problem Set #8: Chapter 4
Answers to Practice Problems
1.
Wendy performs a reaction in which she reacts an organic starting material (C9H15NO)
with bleach to give an organic product (C9H13NO), sodium chloride and water. This
reaction always proceeds with a 52% yield.
(a)
If she has 6.45 g of C9H15NO available for use, what is the minimum volume of bleach
Wendy should use to ensure that all of the organic starting material reacts? (Bleach is a
4.25% w/w solution of sodium hypochlorite in water and has a density of 1.097 g/mL. A
solution described as 4.25% w/w means that 100 g of solution contains 4.25 g of solute.)
Step 1: Find moles of C9H15NO (“SM”)
nSM = mSM =
6.45 g
= 0.0421 mol
USM
153.224 g/mol
Step 2: Find moles of NaOCl
nNaOCl = nSM × mole ratio = 0.0421 mol SM × 1 mol NaOCl = 0.0421 mol NaOCl
1 mol SM
Step 3: Find mass of NaOCl
mNaOCl = nNaOCl × UNaOCl = 0.0421 mol × 74.4419 g/mol = 3.13 g
Step 4: Find mass of bleach solution
mbleach = mNaOCl × 100 g bleach = 3.13 g NaOCl × 100 g bleach = 73.7 g bleach
4.25 g NaOCl
4.25 g NaOCl
Step 5: Find volume of bleach solution
d = m
V
therefore, V = m
d
73.7 g
= 67.2 mL
Vbleach = mbleach =
1.097 g/mL
dbleach
(b)
***3 sig. fig.***
What mass of organic product will Wendy obtain if she does this reaction using 6.45 g of
C9H15NO?
Step 1: Find moles of C9H15NO (“SM”)
6.45 g
= 0.0421 mol
nSM = mSM =
USM
153.224 g/mol
Step 2: Find moles of C9H13NO (“product”)
nproduct = nSM × mole ratio = 0.0421 mol SM × 1 mol product = 0.0421 mol product
1 mol SM
Step 3: Find mass of product (i.e. theoretical yield)
mproduct = nproduct × Uproduct = 0.0421 mol × 151.208 g/mol = 6.37 g
Step 4: Find actual yield
% yield =
actual yield × 100%
theoretical yield
actual yield = % yield × theoretical yield = 52% × 6.37 g = 3.3 g ***2 sig. fig.***
100%
100%
(c)
How much more organic product would Wendy have obtained in part (b) if she were able
to improve the yield of this reaction to 65%?
Repeat Step 4 of part (b) with 65% instead of 52%. This gives an actual yield of 4.1 g.
Subtracting 3.3 g (actual yield for 52%) from 4.1 g (actual yield for 65%) gives 0.8 g.
So, if the reaction proceeded with 65% yield instead of 52% yield, the yield would be
improved by 0.8 g. (1 decimal place therefore 1 sig. fig.)
2.
When lead(IV) oxide is reacted with nitric acid (HNO3), the products are lead(II) nitrate,
oxygen and water.
(a)
Write a balanced chemical equation for this reaction.
2 PbO2 + 4 HNO3
(b)
→ 2 Pb(NO3)2
+
O2 + 2 H2O
If a spoonful of powdered lead(IV) oxide was added to 25.00 mL of 1.416 M nitric acid
and 247.6 mg of oxygen was produced, what was the limiting reactant? (Assume 100%
yield.)
Step 1: Find moles of HNO3
nHNO3 = MHNO3 × VHNO3 = 1.416 mol/L × 0.02500 L = 0.03540 mol
Step 2: Find moles of O2 that would be produced if HNO3 is limiting reactant
nO2 = nHNO3 × mole ratio = 0.03540 mol HNO3 ×
1 mol O2 = 0.008850 mol O2
4 mol HNO3
Step 3: Find moles of O2 actually produced
nO2 = mO2 =
0.2476 g
= 0.007738 mol
UO2
31.9988 g/mol
Step 4: Compare values from steps 2 and 3.
Fewer moles of O2 were produced than would have been if HNO3 was the limiting
reactant. Therefore, something else was the limiting reactant. Since there was only one
other reactant, the limiting reactant must have been PbO2.
3.
Joan wishes to determine the copper content of a copper-containing alloy. After
dissolving 0.251 g of the alloy in acid, she adds an excess of KI(aq). The net ionic
reaction is:
2 Cu2+(aq) + 5 I–(aq) → 2 CuI(s) + I3–(aq)
She then titrates the I3– produced with a sodium thiosulfate (Na2S2O3) solution. The net
ionic reaction is:
I3–(aq) + 2 S2O32–(aq) → S4O62–(aq) + 3 I–(aq)
She finds that 26.32 mL of 0.101 M Na2S2O3 is required for titration to the equivalence
point. Calculate the weight percent of copper in the alloy.
Step 1: Add the two equations to get an overall reaction equation
2 Cu2+(aq) + 2 I–(aq) + 2 S2O32–(aq) → 2 CuI(s) +
S4O62–(aq)
Step 2: Find moles of Na2S2O3
nNa2S2O3 = M Na2S2O3 × V Na2S2O3 = 0.101 mol/L × 0.02632 L = 0.00266 mol
Step 3: Find moles of Cu (assuming moles Cu = moles Cu2+)
nCu = nNa2S2O3 × mole ratio
= 0.00266 mol Na2S2O3 ×
1 mol S2O32- × 2 mol Cu2+ × 1 mol Cu .
1 mol Na2S2O3
2 mol S2O32- 1 mol Cu2+
nCu = 0.00266 mol Cu
Step 4: Find mass of Cu
mCu = nCu × UCu = 0.00266 mol × 63.546 g/mol = 0.169 g
Step 5: Find percent of alloy that is Cu (by mass)
% Cu = mCu × 100% = 0.169 g × 100% = 67.3%
malloy
0.251 g
***3 sig. fig.***
4.
George works in a lab which isolates aluminum metal from aluminum oxide. His step
involves making cryolite (Na3AlF6) by reacting aluminum oxide, sodium hydroxide and
hydrogen fluoride. This reaction also produces water. George’s standard procedure calls
for him to react 85.22 g of aluminum oxide, 176.98 g of sodium hydroxide and 237.51 g
of hydrogen fluoride. Typically, he obtains 306.46 g of cryolite using this procedure.
(a)
Write a balanced chemical equation for this reaction.
Al2O3 + 6 NaOH + 12 HF →
(b)
2 Na3AlF6 + 9 H2O
What is the limiting reactant for George’s procedure?
Step 1: Find moles of each reactant
nAl2O3 = mAl2O3 =
85.22 g
= 0.8358 mol Al2O3
UAl2O3
101.9612 g/mol
nNaOH = mNaOH =
176.98 g
= 4.4248 mol NaOH
UNaOH
39.9971 g/mol
nHF = mHF =
237.51 g
= 11.872 mol HF
UHF
20.0063 g/mol
Step 2: Find n* for each reactant
n*Al2O3 = nAl2O3 = 0.8358 mol Al2O3 = 0.8358 mol Al2O3
1
1
n*NaOH = nNaOH = 4.4248 mol NaOH = 0.73747 mol NaOH
6
6
←LIMITING REACTANT
n*HF = nHF = 11.872 mol HF = 0.98931 mol HF
12
12
(c)
What is the typical percent yield for George’s procedure?
Step 1: Find moles of cryolite using NaOH as limiting reactant
ncryolite = nNaOH × mole ratio = 4.4248 mol NaOH × 2 mol cryolite = 1.4749 mol cryolite
6 mol NaOH
Step 2: Find mass of cryolite (i.e. theoretical yield)
mcryolite = ncryolite × Ucryolite = 1.4749 mol × 209.9413 g/mol = 309.65 g
Step 3: Find % yield
% yield =
5.
actual yield × 100% = 306.46 g × 100% = 98.970 %
theoretical yield
309.65 g
***5 sig. fig.***
Trinitrotoluene (TNT) is made by reacting toluene with nitric acid in the presence of
strong acid catalysts according to the reaction equation below:
CH3C6H5 + 3 HNO3 →
toluene
nitric acid
CH3C6H2(NO2)3
+ 3 H2O
trinitrotoluene
water
Because the each NO2 group is harder to attach than the last, it is necessary to use a large
excess of nitric acid when making TNT. Fortunately, the leftover nitric acid can be
recycled for later reactions.
(a)
If 2.500 L of toluene (density = 0.865 g/mL) is reacted with 15.000 L of concentrated
nitric acid (15.8 M), how many moles of nitric acid will be left at the end of the reaction?
Step 1: Find moles of toluene
mtoluene = Vtoluene × dtoluene = 2500 mL × 0.865 g/mL = 2.16 × 103 g
ntoluene = mtoluene = 2.16 × 103 g = 23.5 mol
Utoluene 92.140 g/mol
Step 2: Find initial moles of nitric acid
nHNO3(i) = MHNO3 × VHNO3 = 15.8 mol/L × 15.000 L = 237 mol
Step 3: Find moles of nitric acid consumed
nHNO3(c) = ntoluene × mole ratio = 23.5 mol toluene × 3 mol HNO3 = 70.4 mol HNO3
1 mol toluene
Step 4: Find moles of nitric acid at end of reaction
nHNO3(e) = nHNO3(i) - nHNO3(c) = 237 mol – 70.4 mol = 167 mol ***0 decimal places***
(b)
TNT explodes releasing 4.184 MJ of energy per kg of TNT. Assuming that the reaction
proceeded with 100% yield, how much energy would be released by detonating the TNT
produced by the reaction described in part (a)?
Step 1: Find moles of TNT
nTNT = ntoluene × mole ratio = 23.5 mol toluene ×
1 mol TNT = 23.5 mol TNT
1 mol toluene
Step 2: Find mass of TNT
mTNT = nTNT × UTNT = 23.5 mol × 227.133 g/mol = 5.33 × 103 g = 5.33 kg
Step 3: Find energy released by this amount of TNT
ETNT = mTNT × 4.184 MJ/kg = 5.33 kg × 4.184 MJ/kg = 22.3 MJ
(c)
Note that burning fats releases 38 MJ of energy per kg and burning sugars releases 17 MJ
of energy per kg. Why is it so much more dangerous to ignite TNT given that it releases
less energy per unit of mass?
TNT releases the energy in one explosive burst whereas burning a fat or sugar releases
the energy more slowly. This makes burning fats or sugars safer than burning TNT.