Chem 115L
Moles, Molarity, Dilution and Stoichiometry
Name________________
Two of the topics from Chem 110 that you must be able to use in Chem 115 are stoichiometry
and solution concentrations/dilutions. The concentration unit most often used in lab is molarity.
In all of the following calculations show work and express the final answer in the correct
number of significant figures.
1. Define molarity of a solution. Be specific with solute, solvent or solution in the definition.
2. One can have a 6.0 M HCl solution
a) only when one has 1 L of solution
b) only when one has 100 mL of solution
c) when one has any volume of solution
3.
a) Formula of copper(II) sulfate pentahydrate is :__________________
b) In what physical state (s, l, or g) will one find the above? ________
c) Molar mass of the hydrated salt (with unit): _________
Show work :
4. Calculate the moles & mass of solute and moles of ions present in each solution:
Moles of
Mass of
Formula
Formula
Moles
Moles
solute
solute
of cation of anion
and M
and M
(mol)
(g)
of
of anion
cation
500.0 mL of
2.00 M
sodium
&
&
thiosulfate
penta_____ _____
hydrate
250.0 mL of
.100 M
ammonium
sulfate
&
&
Total
moles
and M of
ions
&
_____
&
_____ _____
_____
5. What volume of 6.00 M hydrochloric acid will contain 0.100 moles of HCl? _____mL
6. Give the buret reading with units to the correct number of significant figures in the picture below.
Refer to:
http://www.chem.wisc.edu/deptfiles/genchem/lab/labdocs/modules/buret/bretread.htm
24 mL
25 mL
buret reading : ______ unit :_________
7) A sample of 98g of sulfuric acid is dissolved in water to prepare a .50M solution. What is the volume
of the solution made, in liters? __________
8. If 200.00 mL of water was added to the above solution, the new solution’s molarity will be _________
9. To prepare 10. 0L of 0.15 M HCl, one would need _______ mL of 15.0 M HCl and ______ mL of water.
10) If 4.00L of water is added to 1.00 L of a 6.00M solution of calcium chloride, what is the new
molarity of the solution made? Assume additive volumes. ___________
11) 20.0 mL of 0.400 M aq.solution of A is mixed with 60.0 mL of 0.800 M aq. solution of B. If A and B do
not react with each other, the final molarities of the solutions of A = ____ & B = _____ Assume
additive volumes.
12. The molarity of a solution prepared by mixing 300. mL of a 0.250 M solution of sulfuric acid with
300. mL of a 4.00 M solution of sulfuric acid will be _______.
13) 50.0 mL of 3.00 M HCl is mixed with 50.0 mL of 1.00 M NaOH . What is the molarity of the HCl in
the final solution. (Remember there is a reaction between HCl and NaOH)
Setup the ICF Chart after writing the balanced equation: (Use the sample on the last page)
+

+
Initial moles
Which is the
Limiting reactant?
DO NOT USE THESE answers in the division in any subsequent calculations !!!!!!!
They are only to be used to determine the LR.
In all the following steps use initial moles of LR and mole
ratios to find the desired moles
Changed moles
(used or
produced)
Final moles
14. 30.00 mL of 0.200 M FeCl3 solution is reacted with 90.00 mL of 0.100 M NaOH solution.
(a) Write the balanced molecular equation with physical states in paranthesis.
(b) Setup the ICF Chart after writing the balanced equation:
(
)+
(
)
(
)
+
(
)
Initial moles
Which is the
Limiting reactant?
DO NOT USE THESE answers in the division in any subsequent calculations !!!!!!!
They are only to be used to determine the LR.
Changed moles
(used or
produced)
Final moles
Final molarities of
all solutions
(c) From the above chart, mass of the precipitate formed: __________ g
(d) Calculate the total molarities of EACH ION in the resultant solution after reaction.
1. Solution________ : _____M ; cation ______ : ___ M; anion _____: ____ M
2. Solution________ : _____M ; cation ______ : ___ M; anion _____:_____ M
3. Solution________ : _____M ; cation ______ : ___ M; anion _____: _____ M
Total M for each ion:
Ion formula
_________
Molarity
_______
SAMPLE Limiting reactant calculation and ICF Chart
1. Consider the following reaction: NH4NO3 + Na3PO4 (NH4)3PO4 + NaNO3
Which reactant is limiting, assuming we started with 30.0 grams of ammonium nitrate and 50.0 grams of
sodium phosphate. What is the mass of each product that can be formed? What mass of the excess
reactant(s) is left over?
Initial moles
Which is the
Limiting reactant?
3 NH4 NO3
Na 3PO4 
0.3747
0.3049
30.0 g /80.052 (g/mole)
50.0 g/163.94(g/mole)
0.3747/3 = 0.1249
0.3049/1 =0.3049
(NH4 )3 PO4
Zero
3 Na NO3
Zero
Smaller than 0.3049,
Hence
Na 3PO4 is in
excess
DO NOT USE THESE answers in the division in any subsequent calculations !!!!!!!
They are only to be used to determine the LR.
NH4 NO3 is LR
In all the following steps use initial moles of LR and mole
ratios to find the desired moles
Changed moles
(used or
produced)
-0.3747
Final moles
Zero
- 0.1249
(calculate from
initial moles of LR
and the mole
ratio)
0.3049-0.1249=
0.180
(all of initial LR)
+ 0.1249
+0.3747
0.1249
0.3747
Changed moles:
0.3747 moles (LR) NH4 NO3 used x
{
1mol Na 3PO4 used
} = 0.1249 mol Na
3PO4
used
3mole NH4 NO3 used
0.3747 moles (LR) NH4 NO3 used x
{
1mol (NH4 )3 PO4 formed
}
= 0.1249 mol (NH4 )3 PO4 formed
3mole NH4 NO3 used
0.3747 moles (LR) NH4 NO3 used x
{
3mol Na NO3formed
3mole NH4 NO3 used
}
= 0.3747 mol Na NO3formed
Products: (NH4 )3 PO4 : 0.1249 (149.09) = 18.6 g ;
Na NO3 : 0.3747 (84.99) = 31.8 g
Left over excess reactant Na 3PO4 = 0.180 (163.94) = 29.5 g