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-1Chemistry 1010 WInter 2003 #4A
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This is the contents of a Quiz 1 from a few years ago in the days of Chemistry 1000.
(It has been reformatted to save paper)
Answer ALL of the questions in the spaces provided. The mark that you obtain for
this test will be used in calculating your final grade for the course.
1. Name the following compounds:
(b) FeSO4.7H2O iron(II) sulfate heptahydrate
(c) H3PO4(aq)phosphoric acid
(d) K2HPO4(s) potassium monohydrogen phosphate
(e) Cu2O(s) copper(I) oxide
(f) CuH2(s) copper(II) hydride
(a) which reactant, ammonia or oxygen, is the limiting
reagent,
3.00 g
2. Write formulas for the following compounds:
(b) lead(IV) oxide
5. Ammonia gas is oxidized in the following reaction:
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(I)
If 3.00 g of ammonia and 4.00 g of oxygen are allowed
react,
PbO2
mol NH3 = 17.0304 = 0.1762 mol can give 0.1762
mol of NO (1 NH3(g) → 1 NO(g))
4.00 g
(c) nitrogen trichlorideNCl3
(d) sodium dihydrogen phosphate
KH2PO4
(e) ammonium phosphate
(NH4)3PO4
(f) zinc hydroxide
Zn(OH)2
3. (a) Calculate the oxidation number for the atoms
indicated by enlarged symbols in the compounds:
(i) As in As2O3 +3
+4
(ii) Te in TeF5¯
(iii) Br in HBrO2 +3
(b) Identify the oxidizing (oxidant) and the reducing
(reductant) agent in the reaction:
2 Al(s) + 2 OH¯(aq) + 6 H2O(l) → 2 AI(OH)4¯ + 3
H2(g)
Al (O.N. 0) is oxidized to Al3+ (O.N. +3) and Al is the
reducing agent - reductant
H (O.N. +1) in the water is reduced to H2 (O.N. 0) so
the H in the water is the oxidizing agent (oxidant)
4. Write BALANCED NET IONIC EQUATIONS for the
following reactions:
(b) CaBr2(aq) + AgNO3(aq) → AgBr(s) +
Ca(NO3)2(aq)
Br ¯(aq) + Ag+(aq) → AgBr(s)
(c) PbS(s) + HNO3(aq) → Pb(NO3)2(aq) + H2S(g)
PbS(s) + 2 H+(aq) → Pb2+(aq) + H2S(g)
(d) For the reaction between aqueous copper(ll)
sulfate and aqueous sodium hydroxide to give a
blue precipitate.
Cu2+(aq) + 2 OH ¯(aq)→ Cu(OH)2(s)
4
mol O2 = 31.9988 = 0.1250 mol can give 5 x 0.1250
mol of NO = 0.1000 mol of NO
so the O2 is the limiting reagent.
(a) what is the maximum mass (theoretical yield) of
NO(g) that can be formed in the reaction,
From thabove, since the theoretical yield ins moles is
0.05000 mol of NO the mass is
= 0.1000 mol x 30.0061 g mol-1 = 3.000 g
Answer: 3.00 g
(b) and what is the percentage yield if 2.50 g of NO(g)
is produced?
The percentage yield =
Answer: 83.3 %
2.50 g
3.00 g x100%=
83.3 %
6. In the synthesis of ammonia, H2(g) and N2(g) react to
give NH3(g) but the reaction does not go to
completion.
N2(g) + 3 H2(g) → 2 NH3(g)
If the final reaction product mixture from a reaction
contains 2.0 mol of N2(g), 2.0 mol of H2(g) and 2.0 mol
of NH3(g), how many moles of N2(g) and H2(g) were
present at the start of the reaction? (i.e. before any
NH3 formed)?
3
mol H2 used to form NH3 = 2 x 2.0 mol = 3.0 mol
1
mol N2 used to form NH3 = 2 x 2.0 mol = 1.0 mol
So the initial amount is what remains + what was used.
For H2 this = 3.0 mol + 2.0 mol = 5.0 mol
Answer
For N2 this = 1.0 mol + 2.0 mol = 3.0 mol
Answer
7. The following questions refer to acid/base titrations in
which the same solution of base is used.
-2Chemistry 1010 WInter 2003 #4A
(b) What volume of 0.1050 mol L-1 sodium hydroxide
solution, NaOH(aq), is required to neutralize 12.50
mL of 0.1255 moI L-1 hydrochloric acid, HCI(aq)?
The reaction is:
1 NaOH(aq) + 1 HCl(aq) → NaCl(aq) + H2O(l)
9. A sample of carbon monoxide gas, CO(g), occupies
300.0 mL at 27.0 °C and 200.0 kPa pressure.
(b) What volume would the gas occupy at 27.0 °C and
90.0 kPa?
V2 =
-1
moles of HCL used = 0.01250 L x 0.1255 mol L =
1.5688 x 10-3 mol
V2 =
(c) A 0.3200 g sample of an unknown solid diprotic
acid, H2A, requires 24.50 mL of the 0.1050 moI L-1
NaOH solution for neutralization.
2 NaOH(aq) + H2A(aq) → Na2A(aq) + 2 H2O(I)
Calculate the molar mass of the solid acid.
Moles of NaOH used = 0.02450 L x 0.1050 mol L-1 =
2.5725 x 10-3 mol
-3
0.3200 g
and the molar mass of the acid = 1.28625 x 10 −3 mol =
248.79 g mol-1
Ans: 248.8 g mol-1
8. For the reaction
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
propane
(b) 1.0 L (measured at STP) of propane, C3H8(g), is
burned in an excess of oxygen gas what volume of
CO2(g) will be produced at STP?
3.0 L of CO2(g)
P1
P2
x
T2
T1
x V1 =
(c) 500 mL of propane, C3H8(g), is burned in 2.0 L of
oxygen, (both measured at STP) what volume of
CO2(g) would be produced at STP?
C3H8(g) uses 5 O2(g) so 500 mL of C3H8(g) requires
2500 mL oxygen for complete reaction so the O2(g) is
the limiting reagent so
3
Volume of CO2(g) = 5 x 2.0 L = 1.2 L
Ans: 1.2 L
x
273.1 K
300.1 K
x 300.0 mL = 539.0 mL
(d) Calculate the number of mole of CO(g) present in
the sample.
Moles of CO2(g) at STP =
0.5390 L
22.4 L mol −1
= 0.0241 mol
Ans: 0.0241 mol
10.The organic compound called phthalic acid has
analysis:
C 57.83%, H 3.64%, 0 38.52%
and a molar mass of about 166.
(b) Calculate the empirical formula for phthalic acid.
Following the method given in class give the answer,
C4H3O2. (The empirical formula)
Ans: C4H3O2
(c) What is the molecular formula for phthalic acid.
The empirical formula mass is 83.07 while the molar
mass is 166.
166
The empirical formula occurs = 83.07 = 2.00 times in
the molecular formulaso the molecular formula is
C8H6O4
(d) A 6.6115 g sample of a pure liquid was vaporized
in an evacuated 750 mL glass globe at 150.0 °C
and exerted 200 kPa pressure. Calculate the molar
mass of the liquid.
PV
Ans: 3.0 L of CO2(g)
200.0 kPa
101.3 kPa
(Can also use PV=nRT)
Since 2 mole of NaOH reacts with one mole of the acid..
C3H8(g) → 3 CO2(g) so 1.0 L C3H8(g) gives
=666.7 mL
Ans: 539.0 mL
Ans: 14.94 mL
Equal volumes of gases at the same temperature and
pressure contain equal numbers of moles (Avogadro
Hypothesis)
300 mL x 200.0 kPa
90.0 kPa
(c) What volume would the CO(g) occupy at STP?
1,5688 10 −3 mol
amount required is = 0.1050 mol L−1 = 0.014940 L
moles of acid = 2.5725 x 10 x ½ mol = 1.28625 x 10
mol
=
Ans: 667 mL
since the concentration of the NaOH = 0.1050 mol L-1
and one mole is required for each mole of HCl the
-3
P1V1
P2
n = RT
mol
=
200 kPa x 0.750 L
8.314 L kPa K −1 mol −1 x 423.1 K = 0.04264
6.6115 g
so the molar mass is = 0.04264 mol = 155 g mol-1
Ans: 155 g mol-1