Name: __________________________________
Homework 3 –
also see/review the optional HW acids and stuff for additional example problems. Do not turn in acids
and stuff – the key is already up!
1. List the formulas for the elements on the periodic table that naturally exist as diatomic molecules
H2
N2 O2 F2 Cl2 Br2 I2
2. There are 11 elements that naturally exist as gases – what are they – write their name and formula
H2 hydrogen
Cl2 chlorine
Kr krypton
N2
He helium
Xe xenon
O2 oxygen
Ne neon
Rn radon
F2 fluorine
Ar argon
nitrogen
3. There are two elements on the periodic table that naturally exist as liquids – what are they - write
their name and formula
Hg Mercury
Br2 bromine
4. Therefore, all other elements naturally exist as (what phase)___solids______
5. Predict the products of the following single displacement reactions, and provide a balanced
molecular equation based on the descriptions provided. Pay attention to phase labels on this one!
a. __2__Ag(s) + __2___HNO3(aq) 
b. ____Zn(s) + _____CuSO4(aq) 
c. __2___K (s) + ___2__H2O (l)

H2 (g) + 2AgNO3 (aq)
ZnSO4 (aq)
H2 (g)
+ Cu (s)
+ 2KOH (aq)
6. Given the following reactions, predict the products using your solubility rules, write ionic
equations for all and net-ionic if necessary. Name each reactant and product! Do not forget to
show phases!!
a. ____Pb(NO3)4 (aq) + __4__NaCl(aq) →
Names:
PbCl4
+ 4NaNO3 (aq)
(s)
lead (IV) nitrate sodium chloride lead (IV) chloride
Ionic: Pb+4 (aq) + 4NO3-1 (aq) + 4Na+1 (aq) + 4 Cl-1 (aq) → PbCl4
Net-ionic:
+ 4Na+1 (aq) + 4Cl-1 (aq)
Pb+4 (aq) + 4Cl-1 (aq) → PbCl4 (s)
b. ____H2SO4 (aq)
Names:
(s)
sodium nitrate
+ ____Ba(ClO3)2 (aq) →
2HClO3
(aq)
+ BaSO4 (s)
sulfuric acid barium chlorate chloric acid
barium sulfate
Ionic: 2H+1 (aq) + SO4-2 (aq) + Ba+2 (aq) + 2ClO3-1 (aq) → 2H+1 (aq) + 2ClO3-1(aq) + BaSO4 (s)
Net-ionic:
Ba+2 (aq) + SO4-2 (aq) → BaSO4 (s)
c. __2__Li3PO4 (aq) + __3__Cr(C2H3O2)2 (aq) →
6 LiC2H3O2 (aq) + Cr3(PO4)2 (s)
Names: lithium phosphate chromium (II) acetate lithium acetate chromium (II) phosphate
Ionic: 6Li+1 (aq) + 2PO4-3 (aq) + 3Cr+2 (aq) + 6 C2H3O2-1 (aq) → 6Li+1 (aq) + 6 C2H3O2-1 (aq) + Cr3(PO4)2 (s)
Net-ionic:
2PO4-3 (aq) + 3Cr+2 (aq) → Cr3(PO4)2 (s)
7.) Review content check: Potassium iodide is used as a dietary supplement to prevent the iodine
deficiency disease, goiter, and is prepared by the reaction of hydroiodic acid and potassium
hydrogen carbonate. Water, carbon dioxide and potassium iodide are the products of this reaction.
In reacting 245 grams of hydroiodic acid with 116 grams of potassium hydrogen carbonate, how
many grams of potassium iodide are produced? Which reactant is in excess? How many grams of
the excess reagent remain after the limiting reactant has run out (HINT: do NOT use your calculated
amount of product to determine this!! Never ever use calculated numbers to calculate other
amounts unless absolutely necessary!! You have another way to determine this: SUPER DUPER
HINT: Use the LR to determine how many grams of the other reactant would be used up!! – and
then knowing that – how many grams of the excess reactant are left over)? If the reaction had a
82.14% yield, how many grams of potassium iodide were actually made?
First, you need the balanced equation: 1HI + 1KHCO3 → 1H2O + 1CO2 + 1KI
Second: Since information was given to us about BOTH reactants, you need to determine the
LR. The LR is NOT the species that is present in the smallest amount initially – it is the species
that PRODUCES the smallest amount of product. We cannot simply calculate the number of
moles of species and say, ohh the smallest one is always the LR. It just isn’t true!
116 grams KHCO3 x
245 grams HI x
166.0 grams KI
1 mole KHCO 3
1 moleKI
= 192 grams KI (theor)
x
x
100.12 grams HKHCO 3 1 mole KHCO 3
1 mole KI
1 mole HI
1 moleKI 166.0 grams KI
= 318 grams of KI (theor)
x
x
127.9 grams HI 1 mole HI
1 mole KI
So, the KHCO3 gives us one number, and the HI gives us another amount of KI that could be
made. What the smaller amount of KI tells you is that after 192 grams of KI has been made,
there is no more KHCO3 – it is all gone. If there is no more KHCO3, then the HI cannot be
converted into the products at all! So KHCO3 is the limiting reactant and 192 grams is the
theoretical yield!
% yield =
82.14% =
actual amount
x 100
theoretical amount
actual
x 100 =
192
158 grams actually obtained
Finally, the amount of excess reactant that is left over. There are several different ways to solve
this problem. Something to keep in mind. TRY (and you should be able to completely do this!)
to avoid using ANY calculated amounts in this part of the problem. If you have miscalculated
an amount (grams, moles, mL etc. . .) and you use this number to calculate the amount in excess,
your answer will be wrong!! And since you didn’t NEED to use that number, your partial credit
award will be small as well because there was another (aka better) way to solve the problem.
So, if you were told simply that you had 116 grams of KHCO3 and asked how much HI that
would react with, that would be a starting place.
116 grams KHCO3 x
127.9 grams HI
1 mole KHCO 3
1 moleHI
= 148 grams of HI reacted
x
x
100.12 grams HKHCO 3 1 mole KHCO 3
1 mole KI
Well, how much HI did we have? 245 grams to start initially, and of that 245 grams, only 148
grams reacted. So how much is left over?? Exactly – subtract the two from one another!
245 – 148 =
97 grams of HI left over