Chemistry 3A3B 1
Coghlan
1.
Take the reaction: NH
3
+ O
2
NO + H
2
O. In an experiment, 3.25 g of NH
3
are allowed to react with 3.50 g of O
2
. a. Which reactant is the limiting reagent? O
2 b. How many grams of NO are formed? 2.63 g NO c. How much of the excess reactant remains after the reaction? 1.76 g NH
3
left
2.
If 4.95 g of ethylene (C
2
H
4
) are combusted with 3.25 g of oxygen. a. What is the limiting reagent? O
2 b. How many grams of CO
2
are formed? 2.98 g CO
2
3.
Consider the reaction of C
6
H
6
+ Br
2
C
6
H
5
Br + HBr a. What is the theoretical yield of C
6
H
5
Br if 42.1 g of C
6
H
6
react with 73.0 g of Br
2
? 71.8 g
C
6
H
5
Br b. If the actual yield of C
6
H
5
Br is 63.6 g, what is the percent yield? 88.6%
4.
Use the following reaction: C
4
H
9
OH + NaBr + H
2
SO
4
C
4
H
9
Br + NaHSO
4
+ H
2
O If 15.0 g of
C
4
H
9
OH react with 22.4 g of NaBr and 32.7 g of H
2
SO
4
to yield 17.1 g of C
4
H
9
Br, what is the percent yield of this reaction? 61.8%
5.
Silicon nitride (Si
3
N
4
) is made by a combining Si and nitrogen gas (N
2
) at a high temperature.
How much (in g) Si is needed to react with an excess of nitrogen gas to prepare 125 g of silicon nitride if the percent yield of the reaction is 95.0%? 79 g Si
6.
Souring of wine occurs when ethanol is converted to acetic acid by oxygen by the following reaction: C
2
H
5
OH + O
2
CH
3
COOH + H
2
O. A 1.00 L bottle of wine, labelled as 8.5% ethanol, is found to have a defective seal. Analysis of 1.00 mL showed that there were 0.0274 grams of acetic acid in that 1.00 mL. The density of ethanol is 0.816 g mL
-1
and the density of water is 1.00 g mL
-1
. a. What mass of oxygen must have leaked into the bottle? 14.6 g O
2 b. What is the percent yield for the conversion of ethanol to acetic acid if O
2
is in excess? 30.3%
7.
A reaction container holds 5.77 g of P
4
and 5.77 g of O
2
.The following reaction occurs: P
4
+ O
2
P
4
O
6
. If enough oxygen is available then the P
4
O
6
reacts further: P
4
O
6
+ O
2
P
4
O
10
. a. What is the limiting reagent for the formation of P
4
O
10
? O
2 b. What mass of P
4
O
10
is produced? 5.76 g P
4
O
10 c. What mass of excess reactant is left in the reaction container? 5.78 g P
4
O
6
remain

Chemistry 3A3B 2 Coghlan
1. When manganese dioxide reacts with concentrated HCl, the products are manganese (II) chloride, chlorine gas and water. If 2.17 g of MnO
2
are treated with a HCl solution containing 2.74 g of HCl dissolved in 100.0 mL of water, determine:
(a) the limiting reagent
(b) the mass of chlorine produced
(c) the mass of excess reagent left after the reaction.
2. Washing soda is a hydrated form of sodium carbonate. If an 0.374 g sample of washing soda required 25.20 mL of 0.104 mol L
-1
HCl for complete neutralisation, calculate the number of moles of water of crystallization in the sample of washing soda.
3. Hydrochloric acid may be produced by reacting solid sodium chloride with sulphuric acid and passing the HCl gas through water.
(a) Write an equation for this reaction.
(b) Find the volume of HCl gas, collected at STP, which may be produced by completely reacting
0.2925 g of NaCl with sulphuric acid.
(c) What mass of solid Ba(OH) 2 must be added to the solution in (c) to neutralise it?
4. 0.50 g of an alloy of magnesium and aluminium was dissolved in 25.0 mL of 2.00 M HCl. The excess acid required 40.0 mL of 0.10 M NaOH for neutralisation. Calculate the percentage composition of the alloy.
5. A sample of industrial waste water is analyzed and found to contain magnesium sulphate as the only impurity present in a significant amount.
A 5.00 L sample of this waste water was 1 evaporated to about 200 mL and sufficient barium nitrate solution added to precipitate all of the sulphate ion. The precipitate is
Washed with distilled water, filtered and dried. The mass of the precipitate is found to be 1.70 g.
( a) Determine the concentration in mg L
-1
of magnesium sulphate in the waste water.
(b) What mass of washing soda, Na
2
CO
3
.10H
2
O, would need to be added to 100.0 L of the waste water to cause the precipitation of all the magnesium ions?
6. Potassium perchlorate (KClO4) can be prepared by the following series of reactions:
Cl2 + 2 KOH

KCl + KClO + H20
3KClO

2KCl + KClO3
4KClO3

3KClO4 + KCl
What mass of chlorine is required to produce 50.0 g of KClO4?

Chemistry 3A3B 3 Coghlan
1. MnO2(s) + 4HCl(l)
(a) n (MnO2)
= 2.17/86.94
= 0.0250 mol
MnCl2(aq) + Cl2(g) + 2H2 O(l)
n (HCl) = 2.74/36.46
= 0.0752 mol
If 0.0250 mol MnO2 are present, 0.100 mol HCl are required - therefore HCl is the limiting reagent.
(b) n (Cl2) = 0.0752/4
m (Cl2) = (0.0752/4) x 70.90
= 1.33 g
(c) n (MnO2) used in the reaction
= 0.0752/4 mol n (MnO2) left
= 0.0250 - (0.0752/4)
= 0.00620 mol m (MnO2) = 0.00620 x 86.94
= 0.539 g
2. Na2CO3(s) + 2HCl(aq) n (HCl)
2NaCl(aq) + CO2(g) + H2O(l)
= (25.20/1000) x 0.104 n (Na2CO3) = 1/2 x (25.20/1000) x 0.104 m (Na2CO3) = 1/2 x (25.20/1000) x 0.104 x 105.99
= 0.139 g m (H2O) = 0.374 - 0.139
= 0.235 g n (H2O) = 0.235/18.016
= 0.0130 mol n (Na2CO3) = 0.00131 mol
Na2CO3.10 H2O
3. (a) 2NaCl(s) + H2SO4(aq) Na2SO4(aq) + 2HCl(g)
NaCl(s) + H+(aq) Na+(aq) + HCl(g)
(b) n (NaCl) = 0.2925/58.44
n (HCl)
V (HCl)
= 5.005 x 10-3 mol
= 5.005 x 10-3 mol
= 5.005 x 10-3 x 22.4
=0.112 L
(c) 2HCl(aq) + Ba(OH)2(aq) n (HCl)
BaCl2(aq) + 2H2O(l)
= 5.005 x 10-3 mol
= 1/2 x 5.005 x 10-3 mol n (Ba(OH)2) m (Ba(OH)2) = 1/2 x 5.005 x 10-3 x 171.33
= 0.429 g

Chemistry 3A3B 4
4. n (HCl) = (25/1000) x 2
= 0.0500 mol n (NaOH) extra = (40/1000) x 0.1
= 0.00400 mol
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) n (HCl) used = 0.05 - 0.004
= 0.046 mol
Only the magnesium will react with the HCl
2HCl(aq) + Mg(s) n (Mg)
MgCl2(aq) + H2(g)
= 1/2 x 0.046 m (Mg) = 1/2 x 0.046 x 24.31
= 0.56 g
% Mg m(Al)
% Al
= 0.56/0.70 x 100
= 80.0 %
= 0.14 g
= 0.14/0.70 x 10
= 20.0 %
5. SO42-(aq) + Ba2+(aq)
(a) n (BaSO4)
BaSO4(s)
= 1.70/233.4 n (MgSO4)
= 7.28 x 10-3 mol
= 7.28 x 10-3 mol c (MgSO4) = 7.28 x 10-3 /5.00
= 1.46 x 10-3 mol L-1 = 175 mg L-1
(b) n (MgSO4) in 100 L of waste water
= 7.28 x 10-3 x (100/5.00)
CO32-(aq) + Ba2+(aq) BaCO3(s) n (Na2CO3.10H2O) = 7.28 x 10-3 x (100/5.00) m (Na2CO3.10H2O) = 7.28 x 10-3 x (100/5.00) x 286.15
= 41.7 g
6. From the balanced equations
12 moles of Cl
2
produce 3 moles of KClO
4 n(KClO
4
) = 50.0/138.55 n(Cl
2
) m(Cl
2
)
= 50.0/138.55 x 4
= 50.0/138.55 x 4 x 70.9
= 102 g
Coghlan