Ch 60. Exam 2. Sample key
Info: 6.022 x 1023 atoms = 1 mole also 1 g = 6.022 x 1023
1. The molar mass of an oxygenated hydrocarbon substance is 192.12 g / mole . Its composition is as
follows: C : 37.5 % , H : 4.2 % , O : 58.29 % Calculate the empirical formula and
molecular formula. (use atomic masses C = 12.0 , H = 1.0 , O = 16.0 )
37.5 g C / 12.0 = 3.125 mole / 3.125 = 1.00 x 6 = 6
4.2 g H / 1.008 = 4.17 mole / 3.125 = 1.33 x 6 = 8
58.29 g O/16.00 = 3.64 mole / 3.125 = 1.16 x 6 = 7
so: C6H8O7 m mass = 192 same as empirical
2. How many grams of bromine are contained in 7.50 g of ammonium bromide ?
M mass = 79.9 + 14.0 + 4.03 = 97.93
7.50 g NH4Br x 79.9 = 6.12 g
97.93
.
3. Calculate the mass, in grams of 2.06 x 1024 boron atoms / 6.02 x 1023 = 3.42 mole x 10.8 g/mole = 37.3 g
4. Name the following compounds according to the IUPAC system
a) Pb(IO3)4 lead(IV) iodate
b) HCN (aq) hydrocyanic acid
c) K3Sb potassium antimonide
d) CsH cesium hydride e) H3PO3 phsphorous acid f) C4H10 butane g) AsCl5 arsenic pentachloride
h) Sn(C2H3O2)2 tin(II)acetate
5. Write formulas for the following named compounds:
a) copper(I) carbonate Cu2CO3
b) xenon hexafluoride XeF6
c) heptane C7H16
d) cobalt(III)perbromate Co(BrO4)3
e) calcium peroxide _CaO2
f) silicic acid H2SiO3
g) lithium cyanide -LiCN h) potassium hydrogen carbonate KHCO3
8. Write balanced equations for the following reactions . Assume they do give products.
a) calcium metal reacts with water to produce a gas and a compound in solution
Ca + H2O  H2 + CaO
Ca + 2 H2O  H2 + Ca(OH)2
CaO + H2O  Ca(OH)2
b) barium hydroxide reacts with bromic acid
Ba(OH)2 + 2 HBrO3  Ba(BrO3)2 + 2 H2O
c) manganese (III) oxide combines with water to form one substance Mn2O3 + 3 H2O  2 Mn(OH)3
d) dinitrogen trioxide reacts with water to produce one substance (an acid) N2O3 + H2O  2 HNO2
e) pentane combusts
C5H12 + 8 O2 
f) sulfuric acid decomposes
5 CO2 + 6 H2O
H2SO4  SO3 + H2O
g) H2 reacts with Fe3O4 to produce an element and a compound
4 H2 + Fe3O4  3 Fe + 4 H2O
h) hydrogen and carbon monoxide can be produced by reacting coal with steam
C + H2O  CO + H2
i). phosphoric acid reacts with magnesium hydroxide, forming two products
2 H3PO4 + 3 Mg(OH)2  Mg3(PO4)2 + 6 H2O
9. 10.0 grams of potassium carbonate is mixed with 10.0 grams of hydrochloric acid . 1.25 L of
a gaseous product was collected at STP
 CO2 + H2O + 2 KCl
a) Write the balanced equation K2CO3 + 2 HCl
b) Calculate the moles of the two reactants and determine which is the limiting reactant.
10.0 g K2CO3 / 138.2 g/mole = 0.0724 mole
10.0 g HCl / 36.5 g/mole = 0.274 mole
X 2 HCl = 0.1448 moles HCl needed to react with 0.0724 moles K2CO3
0.274 moles HCl given – so too much HCl
K2CO3 is the limiting reactant
c) Calculate the theoretical yield and the percent yield in the reaction
0.0724 mole K2CO3 x 1 mole CO2 / 1 mole K2CO3
x 22.4 L CO2 / mole CO2 = 1.62 L CO2
(theo yield)
Actual yield ( 1.25 / theo 1.62 ) x 100 = 77.2 %
16. For the reaction between barium iodide (aq) and potassium phosphate (aq) ,
a) write the total balanced equation . indicate which product precipitates (s)
3 BaI2 + 2 K3PO4 
Ba3(PO4)2 + 6 KI
b) write a total ionic equation, and indicate which are the spectator ions ( SI )
3 Ba2+ + 6 I- + 6 K+ + 2 PO43-  Ba3(PO4)2 (s) + 6 K+ + 6 Ic) write the net ionic equation
3 Ba2+ + 2 PO43-  Ba3(PO4)2 (s)
17. Using the reaction in problem 16, predict how many grams of product would precipitate after mixing two
…, solutions of reactants.
Solution 1 contained 10.0 g of barium iodide.
Solution 2 contained 8.00 grams of potassium phosphate.
Show all work – and indicate which reactant is the limiting reactant.
lim. reactant 5 pts
mol. mass calcs 5 pts
mole calcs 10 pts incl 2 pts sig sigs
product
15 pts incl 2 pts sig figs
Atoms &
Molecules
.
x by atomic mass
.
/. by Avogadro’s #
/.
x by Avog. #
MOLES
amu
.
.
/. by molar mass
/. by Avog #
x by Avog #
x by molar mass
GRAMS
Atoms &
Molecules
.
x by atomic mass
.
/. by Avogadro’s #
/.
x by Avog. #
MOLES
amu
.
.
/. by molar mass
/. by Avog #
x by Avog #
x by molar mass
GRAMS