Chemistry 350
20 points
Spring 2016
Due In Class Wednesday January 20
Problem Set 1 -- Stoichiometry and Concentration Review
Complete all problems on separate paper. Show all work for credit. Correct use of
significant figures is required for full credit.
1. Describe how to prepare 2.00 L of a solution that has a chloride concentration of
0.0100 M starting with:
a. solid copper (II) chloride
b. 0.200 M copper (II) chloride solution.
a.
0.010 mol Cl- x 2.00 L x 1 mol CuCl2 x 134.452 g CuCl2 = 1.35 g CuCl2
1L
2 mol Cl1 mol CuCl2
So, weigh 1.35 g CuCl2 and dilute it to 2.00 L.
b.
1L
= 0.050 L CuCl2
0.010 mol Cl- x 2.00 L x 1 mol CuCl2 x
1L
2 mol Cl0.200 mol CuCl2
So, diluted 0.050 L (50 mL) of 0.200 M CuCl2 ti 2.00 L.
2. When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon
dioxide, and water are produced. How many grams of calcium chloride will be
produced when 31.0 g of calcium carbonate are combined with 10.0 g of
hydrochloric acid?
CaCO3 + 2HCl → H2O + CO2 + CaCl2
31.0 g CaCO3 x
10.0 g HCl x
1 mol CaCO3 x 1 mol CaCl2 x 110.984 g CaCl2 = 34.37 g CaCl2
100.067 g CaCO3 1 mol CaCO3
1 mol CaCl2
1 mol HCl x 1 mol CaCl2 x 110.984 g CaCl2 = 15.22 g CaCl2
2 mol HCl
1 mol CaCl2
36.461 g HCl
So, HCl is the limiting reactant and 15.2 g of CaCl2 will be produced.
3. 0.850 L of 0.490 M sulfuric acid is mixed with 0.800 L of 0.250 M potassium
hydroxide. What concentration of sulfuric acid remains after neutralization?
H2SO4 + 2 KOH → 2H2O + K2SO4
0.800 L KOH x 0.250 mol KOH x 1 mol H2SO4 = 0.100 mol H2SO4 consumed
2 mol KOH
1 L KOH
0.850 L H2SO4 x 0.490 mol H2SO4 = 0.4165 mol H2SO4 introduced
1 L H2SO4
0.4165-0.100 = 0.3165 mol H2SO4 remain/(0.800+0.850)L = 0.192 M H2SO4
Chemistry 260
Chapter 1 – Introduction
2
4. In the laboratory, you weigh out 0.114 grams of solid AlCl3•6H2O (molar mass
241.43 g/mol), dissolve it, and dilute it to a total volume of 100.0 mL to make
solution A. You then transfer 3.00 mL of solution A into a 25.0 mL volumetric flask
and dilute to the mark to make solution B. What is the concentration of aluminum
ion in solution B in moles per liter? In ppm?
Solution A:
0.114 g AlCl3•6H2O x
Solution B:
In ppm:
1 mol AlCl3•6H2O x
1 mol Al3+
x
1
= 4.72x10-3 M
241.43 g AlCl3•6H2O 1 mol AlCl3•6H2O 0.1000 L
3 mL x 4.72x10-3 mol Al3+ x 1 = 5.67x10-4 M
1L
25 mL
5.67x10-4 mol Al3+ x 26.982 g Al x 103 mg = 15.3 mg Al = 15.3 ppm
1g
1L
1L
1 mol
5. An iron ore sample weighing 0.9132 g is dissolved in HCl(aq), and the iron is
obtained as Fe2+(aq). This solution is then titrated with 28.72 mL of 0.05051 M
K2Cr2O7(aq) using the balanced reaction below. What is the mass percent of iron in
the iron ore?
6Fe2+ + 14H+ + Cr2O72- → 6Fe3+ + 2Cr3+ + 7H2O
0.0287 L x 0.05051 mol Cr2O72- x 6 mol Fe x 55.845 g Fe = 0.4857 g Fe
1 mol Cr2O721 mol Fe
L
0.4857 g Fe x 100% = 53.19% Fe
0.9132 g ore