Chemistry 2 AP/AC
Unit 2 HW 2
Name_______________________________
1) Characterize the following ionic compounds as soluble (S) or insoluble (I) in water.
(a) Ca3(PO4)2 _____
(b) Mn(OH)2 _____
(c) AgClO3 _____
(d) K2S _____
(e) CaCO3 _____
(f) ZnSO4 _____
(g) Hg2(NO3)2 _____
(h) NH4ClO4 _____
(i) PbCl2 _____
2) Determine and write the products of the following precipitation reactions. Then write
the ionic and net ionic equations. Make sure that the net ionic equation is balanced.
(a)
AgNO3(aq) + KI(aq) 
(b)
BaCl2(aq) + ZnSO4(aq) 
(c)
(NH4)2CO3(aq) + CaCl2(aq) 
(d)
Na2S(aq) + ZnCl2(aq) 
(e)
K3PO4 (aq) + Sr(NO3)2 (aq) 
3) How many mL of 1.50 M HCl would be necessary to dissolve 1.37 grams of
Mg according to the reaction below?
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
4) Suppose we place 3.50 grams of MgCl2 into a flask and add 275 mL of
a 2.00 M AgNO3 solution. What is the limiting reagent and how many grams
of AgCl will precipitate?
MgCl2 (s) + 2AgNO3 (aq) 
2AgCl (s) + Mg(NO3)2
5) Limestone (CaCO3) reacts with hydrochloric acid to produce water,
an aqueous solution of calcium chloride, and carbon dioxide gas.
Suppose 50.0 mL of 0.650 M HCl is added to 2.25 grams of limestone.
a) Write the balanced molecular equation.
b) Calculate how many grams of CO2 gas will be released.
6) What mass of barium sulfate is produced when 100.0 mL of 0.100 M barium chloride
solution is mixed with 100.0 mL of 0.100 M solution of iron(III) sulfate? It may help
to first write a balanced reaction.
7) Page 182, #43, parts (a) – (d)
8) Chlorisondamine (C14H18Cl6N2) is a drug used in the treatment of high blood pressure.
A 1.28 g sample of an impure sample of the drug was dissolved in a solution that
released all of the chlorine as chloride ion. The solution was then treated with excess
silver nitrate, and silver chloride precipitated. The AgCl precipitate was filtered, dried
and weighed. The amount of AgCl collected was 0.104 g. From this data, calculate
the mass percent of chlorisondamine in the medication.
Answers to Unit 2 HW 2
1)
I
S
S
2)
(a)
(b)
(c)
(d)
(e)
I
I
S
net ionic:
net ionic:
net ionic:
net ionic:
net ionic:
S
S
I
Ag+(aq) + I-(aq) → AgI(s)
Ba+2(aq) + SO4-2(aq) → BaSO4(s)
Ca+2(aq) + CO3-2(aq) → CaCO3(s)
Zn+2(aq) + S-2(aq) → ZnS(s) b
3Sr+2(aq) + 2PO4-3(aq) → Sr3(PO4)2(s)
3)
1.37 g Mg · 1 mol Mg · 2 mol HCl ·
1L
· 1000 mL HCl
24.31 g
1 mol Mg 1.50 mol HCl
1L
= 75.1 mL HCl
4)
3.50 g MgCl2 · 1 mol MgCl2 · 2 mol AgNO3 · 1 L AgNO3
· 1000 mL
95.21 g
1 mol MgCl2 2.00 mol AgNO3
1L
= 36.8 mL AgNO3 needed
MgCl2 = LR
3.50 g MgCl2 · 1 mol MgCl2 · 2 mol AgCl · 143.32 g AgCl
95.21 g
1 mol MgCl2 1 mol AgCl
5)
(a)
= 10.5 g AgCl
CaCO3(s) + 2HCl(aq) → H2O(l) + CaCl2(aq) + CO2(g)
(b) 50.0 mL HCl → → → 1.63 g CaCO3 needed
HCl = LR
50.0 mL HCl → → → 0.715 g CO2 produced
6)
3BaCl2(aq) + Fe2(SO4)3(aq) → 2FeCl3(aq) + 3BaSO4(aq)
0.1000 L BaCl2 · 0.100 mol BaCl2 · 1 mol Fe2(SO4)3 ·
1L
·
1L
3 mol BaCl2
0.100 mol Fe2(SO4)3
1000 mL
1L
= 33.3 mL Fe2(SO4)3 needed
BaCl2 = LR
0.1000 L BaCl2 · 0.100 mol BaCl2 · 3 mol BaSO4 · 233.40 g BaSO4 = 2.33 g BaSO4
1L
3 mol BaCl2
1 mol BaSO4
7)
(a) 2KOH(aq) + Mg(NO3)2(aq) → Mg(OH)2(s) + 2KNO3(aq)
(b) Mg(OH)2
(c) 100.0 mL KOH → → → 50.0 mL Mg(NO3)2 needed so KOH = LR
100.0 mL KOH → → → 0.583 g Mg(OH)2 produced
(d) [K+] = [OH-] = (0.1000 L) (0.200 mol/L) = 0.100 M
(0.2000 L)
Similarly, [Mg+2] = 0.100 M and [NO3-] = 0.200 M
Initial [ ]
Change in [ ]
Final [ ]
Mg+2 + 2OH− → Mg(OH)2
0.100 M
0.100 M
-0.050 M -0.100 M
0.050 M
0M
So after the reaction occurs:
[K+] = 0.100 M
[OH-] = 0 M
[Mg+2] = 0.050 M
[NO3-] = 0.200 M
8)
% Cl = 212.70 g X 100% = 49.808 %
427.04 g
% Cl = 35.45 g X 100% = 24.73 %
143.32 g
g Cl = (0.2473) (0.104 g) = 0.0257 g
Let x = g of C14H18Cl6N2
0.0257 g Cl = (0.49808) (x)
x = 0.0516 g
% purity = 0.0516 g X 100% = 4.03 %
1.28 g