Download CHEM 115 EXAM #2

CHEM 115 EXAM #2
PRACTICE
Solve the following problems. Show all your work, always show your units, and express
your answers in the proper number of significant figures. (point values as indicated)
(10 points)
1. Iron (II) chloride (FeCl2) reacts with silver nitrate (AgNO3) in aqueous solution to
form a precipitate. Write the balanced molecular (total), the balanced ionic, and
the balanced net ionic equations for this reaction.
molecular
full ionic Fe
2+
net ionic 2 Cl
FeCl2 (aq) + 2 AgNO3
-
(aq)
+ 2 Cl
(aq)
+ 2 Ag (aq) Æ 2 AgCl (s)
-
(aq)
(aq)
Æ 2 AgCl (s) + Fe(NO3)2
+ 2 Ag (aq) + 2NO3
or
(aq)
(aq)
Æ 2 AgCl (s) + Fe2+(aq) + 2NO3
Cl
(aq)
(aq)
+ Ag (aq) Æ AgCl (s)
(15 points)
2. (a) Balance the following reaction.
6 HCl (aq) + 2 Fe (s)
→ 2 FeCl3 (aq) + 3 H2 (g)
If 5.14 g of Fe is reacted with 50.0 mL of 6.00 M HCl:
(b) What is the limiting reactant?
determine moles of Fe and HCl and compare to find the LR is Fe(s) b/c
5.14 g
= 0.09203moleFe
g
55.85
mole
⎛ 6.00mole ⎞
moleCl2 = ⎜
⎟ * 0.0500 L = 0.300 0 moleHCl
L
⎝
⎠
moleFe =
logic. . .at a mole ratio of 6:2, 0.276 mole of HCl is needed to react with 0.920 mole of
Fe, since we have 0.300 mole of HCl it is present in a very small excess
(c) What is the theoretical yield of H2 in both moles and grams?
convert mole of Fe to mole of H2 gas and then into grams of H2 gas
3 mole H2 for every 2 mole of Fe yields 0.138 mole of H2
with each mole of H2 having a mass of 2.016 g to give 0.278 g H2
(d) What amount of H2 would be produced if the percentage yield for the reaction
is 89.0%? Provide your answer in both moles and grams.
0.890*(answers in part c) = 0.123 mole H2 = 0.248 g H2
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(10 points)
5. a. How many moles of NaCl are required to make 2.50 x 102 mL (250 mL) of a
solution that has a concentration of 1.50 M?
mole = (
1.50 mole
/L sol’n)(0.250 L) = 0.375 mole
b. What mass of NaCl is required to prepare the same solution?
mass = (
58.44 g
/1 mole solid)(0.375 mole) = 21.92 g
(15 points total, 2.5 points each)
6. Complete and balance the common reactions listed below (include physical state
labels).
a. metal + nonmetal → binary salt compound
Mg (s) +
Cl2 (g) →
MgCl2 (s)
b. metal oxide + nonmetal oxide → salt of the metal ions and oxoanions of the
nonmetal
CaO (s) +
CO2 (g) →
CaCO3 (s)
c. metal oxides (basic oxides) + water → metal hydroxide
CaO (s) +
H2O (l) →
Ca(OH)2 (aq)
d. nonmetal oxides (acidic oxides) + water → oxoacid
SO3 (s) +
H2O (l) →
H2SO4 (aq)
e. hydrocarbon + oxygen → carbon dioxide + water
C3H8 (s) +
5 O2 (g) →
3 CO2 (g)
+
4 H2O (g)
+
H2 (g)
f. active metal + acid → salt + hydrogen gas
Mg (s) +
H2SO4 (aq) →
MgSO4 (aq)
g. active metal + salt of less active metal → salt of active metal + less active
metal
Zn (s) +
CuSO4 (aq) →
Cu (s) +
114
ZnSO4 (aq)
(10 points)
7. H2SO4
or HCl
Ba(OH)2
NH3
HF
a. Select ONE strong acid and write a molecular reaction (balanced) that shows
what happens when the strong acid is added to water.
H2SO4
or
HCl
H2SO4 (aq)
or
HCl (aq)
+
H2O (l) → H3O+(aq)
+
HSO4
+
H2O (l) → H3O+(aq)
+
Cl
(aq)
(aq)
b. Select ONE weak base and write a molecular reaction (balanced) that shows
what happens when the weak base is added to water.
NH3
NH3 (aq)
+
-
H2O (l) → OH (aq)
+
NH4
+
(aq)
c. Write the balanced molecular reaction that would occur between HCl (aq) and
Ba(OH)2 (aq).
Ba(OH)2 (aq) + 2 HCl (aq) → 2 H2O(l) +
BaCl2 (aq)
(15 points)
8. a) Balance the following redox equation in acidic solution (SHOW ALL
WORK!!!!!):
(unbalanced)
(oxid)
+ IO4-(aq) + Fe2+(aq) → I-(aq) + Fe3+(aq) + H2O(l)
H+(aq)
8H
+
(aq)
+
8 e-
+ IO4-(aq) → I-(aq) + 4 H2O(l)
(red) [ Fe2+(aq) → Fe3+(aq) + e- ]x8
8H
+
(aq)
+
8 Fe2+(aq) + IO4-(aq) → I-(aq) + 8 Fe3+(aq) + 4 H2O(l)
b) The reaction above is carried out. Determine the molarity of the Fe2+ solution
if it required 44.76 mL of 0.0194 M IO4- solution to completely neutralize 25.00
mL of the Fe2+ solution.
MFe
2+
8moleFe 2+ 0.0194 moleIO4− 0.04476 mLIO4− sol ' n
x
x
=
= 0.277 9 M
1moleIO4−
1LIO4− sol ' n
0.02500 mLFe 2+ sol ' n
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9.
(10 points)
The electrolyte in a lead storage battery is dilute sulfuric acid, H2SO4 (aq). This
acid must have a concentration between 4.8 M and 5.3 M if the battery is to be
most effective. A 5.00 mL sample of battery acid [H2SO4 (aq)] is completely
neutralized by 46.40 mL of 0.875 M sodium hydroxide [NaOH (aq)]. Balance
the chemical reaction below and calculate the molarity of the original 5.00 mL
H2SO4 (aq) sample.
H2SO4 (aq) +
Macid =
2 NaOH (aq)
→
Na2SO4 (aq)
+
2 H2O (l)
1moleH 2 SO4 0.875moleNaOH 0.04640mLNaOHsol ' n
x
x
= 4.06 M
2moleNaOH
1LNaOHsol ' n
0.00500mLH 2 SO4 sol ' n
The acid is not within the optimal range for peak battery performance!
(15 points)
10.
An aqueous solution of Na3PO4 has a concentration of 2.50 x 10-5 M. Find the
following:
a. What is the concentration of sodium ions in this solution?
There are three sodium ions per formula unit, so [Na+] = 3*(2.50 x 10-5 M)= 7.50 x 10-5M
b. What is the concentration of phosphate ion in this solution?
There is one PO43- ion per formula unit, so [PO43-] = 1*(2.50 x 10-5 M)= 2.50 x 10-5M
c. What is the minimum volume (in mL) of 0.125 M FeCl3 solution needed to
cause the precipitation of all of the phosphate present in 250 mL of the sodium
phosphate solution?
Fe3+(aq) + PO43-(aq) → FePO4 (s)
1000mLFeCl3
1moleFeCl3 2.50 x10 −5 moleNa3 PO4
Vol =
x
x
x 250mLNa3 PO4 sol ' n
0.125moleFeCl3 1moleNa3 PO4 1000mLNa3 PO4 sol ' n
Vol = 0.050mLFeCl3 sol ' n
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BONUS (for up to five, 5, points answer the following correctly)
For the solution of sodium phosphate in the previous question, 2.50 x 10-5 M Na3PO4:
How many sodium ions are present in each µL of the solution? note: 1 µL = 1x10-6 L
Na + ions =
2.50 x10 −5 moleNa3 PO4
6.022 x10 23 Na +
3moleNa +
x
x1x10 −6 L
x
+
1moleNa3 PO4
1LSol ' n
1moleNa
Na + ions = 4.5 x10 +13 Na + ions
How many phosphate ions are present in each µL of the solution?
For every three Na+ there is only one phosphate so . . .
1.5x10+13 PO43- ions will be found in each µL of sol’n
What is the TOTAL ion concentration present in units of ions per µL?
The total is the sum of both the sodium and the phosphate
total = 6.0x10+13 ions in each µL of sol’n
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