Fall, 2000 Name: __________________ Third ExamExam begins here. There are 30 questions.
Questions 1-4 are worth 2 pts each.
Three types of equations are used to describe reactions in aqueous solutions.
1. The molecular equation shows the overall reaction but not necessarily the actual forms of the reactants and products in solution.
2. The complete ionic equation represents all reactants and products that are strong electrolytes (completely dissociate) as ions. All reactants and products are included.
3. The net ionic equation includes only those components that undergo a change. Spectator ions are not included.
4. What subatomic particle is transferred during an oxidation- reduction reaction?
Write the complete molecular, complete ionic, and net ionic equations for each of the following reaction equations. Show spectator ions and cross them through (italicized here) to indicate that they are spectator ions.
5.
H+(aq) + Cl-(aq) + Rb+(aq) + OH-(aq) ---> Rb+(aq) + Cl-(aq) + H2O(l)
H+(aq) + OH-(aq) ---> H2O(l)
These 3 questions are worth 2 points each.
7. H2O2(aq) ---> H20(l) + O2(g)
oxidation-reduction
8. AgNO3(aq) + CuCl2(aq) ---> Cu(NO3)2 + AgCl(s)
precipitation
9. Ca(OH)2(aq) + HNO3(aq) ---> Ca(NO3)2(aq) + H2O(l)
acid-base
For each of the following metals, how many electrons will the metal lose when it ionizes (forms its ion)?
The next 6 questions are worth 1 point each.
10. sodium 1
11. barium 2
12. aluminum 3
For each of the following nonmetals, how many electrons will the nonmetal gain when it ionizes?
13. nitrogen 3
14. sulfur 2
15. chlorine 1
The next 6 questions rely upon your knowledge of Avogadro's Number, moles, molecules, and atoms. REQUIRED: Show your calculations.
These problems are worth 3 points each.
16. How many atoms are there in 2 moles of He?
1 mole contains 6.022 x 1023 atoms so 2 moles must contain 2 x 6.022 x 1023 which is 12.04 x 1023, and in scientific notation this is 1.204 x 1023.
17. Calculate the mass of one atom of lithium (Li). Show the mass both in a.m.u. and in grams (g).
The atomic mass on the Periodic Table is a.m.u.
for a single atom so the mass of Li in a.m.u. is 6.941.
The mass of 6.941 is grams when discussing a mole of Li. A mole
contains 6.022 x 1023 atoms. The mass of a single
atom would have units of g/atom. So one needs to divide the mass
in grams by the number of atoms.
-------------------
6.022 x 1023 atoms/mole
18. A sample of "stuff" is found to weigh 54 grams. The "stuff" contains 2 hydrogen atoms for each oxygen atom in the compound. The simplest formula shows 2 hydrogens to 1 oxygen. How many moles of this "stuff" is in the 54 grams?
The simplest formula shows that the compound must be
H2O. THe molar mass of H2O is 18 g.
Divide the total number of grams by the molar mass and one gets a
value of 3 moles.
The mass of one mole of CaCO3 is 100 g so 0.5 mole must equal 50 g.
First calculate the molar mass of NaOH which is: 23 + 16 + 1 = 40 g.
Next set up a ratio;
1 mole X moles
------ = ---------
40 g 100 g
The answer is: 2.5 moles
1 mole of water (or any substance for that matter) always
contains 6.022 x 1023 molecules
Set up a ratio:
1 mole 2.62 x 10-6 moles
----------- = ------------------
6.022 x 1023 X molecules
The answer is: 1.58 x 1018 molecules
These two problems are worth 6 points each.
22. A 10.00 g sample of copper metal is heated strongly in air. After reacting, the sample weighs 12.52 g. The additional weigh comes from oxygen which reacted with the copper metal. Determine the molecular formula using the % composition method.
The compound is made of Cu and O. According to the Law of
Conservation of Matter the mass of Cu going in equals the mass of
Cu coming out so, 10.00 g of Cu come out and the difference of
2.52 g is oxygen.
Next calculate the number of moles of Cu and O.
10.00 g
----------- = 0.157 mole
63.5 g/mole
2.52 g
----------- = 0.157 mole
16 g/mole
So the ratio is 0.157 to 0.157 which is equivalent to 1:1 and the
formula therefore is CuO.
The empirical formula is a reduced form of a formula where
common factors for each atom are divided out. The molecular
formula is the actual formula for the molecule being studied.
The empirical formula has a mass of 13. The molar mass of the
molecular compound is 78 so divide 78 by 13 to obtain the number
of CH units in the molecular compound. 78/13 = 6 so there must
have been a common coefficient of 6 for the C and the H. The
molecular formula is C6H6.
24. (4 pts) What is the limiting reactant for a chemical reaction? Explain.
The limiting reactant is the reactant in the reaction which is
less than the balanced ratio of moles needed to react with the
other reactant(s).
The reaction will continue until the limiting reactant is used up
then the reaction will stop.
The theoretical yield is the yield calculated using the balanced
chemical equation as a guide. This is the amount of product that
should be produced.
The actual yield is the amount of product recovered after an
experiment is conducted. As a rule the actual yield is less than
but can equal the theoretical yield.
1.279 g
% yield = --------- x 100 = 94.6 %
1.352 g
27. (6 pts) When glucose, C6H12O6, is burned completely in air,
CO2 and water are produced. Write a balanced chemical equation
for this reaction. Then calculate the theoretical yield of CO2
is 1.00 g of glucose is burned completely.
First the balanced chemical equation must be determined so that
the number of moles of CO2 produced per mole of
glucose can be determined. Actually that is not necessary in
this reaction since all carbon coming into the reaction ends up
in a single product. Glucose contains 6 moles of carbon atoms in
each mole of glucose so the conplete combustion will produce 6
moles of CO2 per mole of glucose.
Next find the number of moles equivalent to 1 g of glucose. The molar mass is 180 g/mole. So:
6 x 5.56 x 10-3 moles of CO2 are produced
from 5.56 x 10-3 moles of glucose. This equals 3.33 x
10-2 moles of CO2.
180 g 1.00 g
--------- = -----------
1 mole x moles
X = 5.56 x 10-3 moles of glucose.
A) 0.000305 mole mercury (Hg) 6.12 x 10-2 g
B) 125 moles ammonia (NH3) 2.125 x 103 g
C) 0.01205 mole of sodium peroxide (Na2O2) 0.94 g
29. (6 pts) Using :
First balance the equation.
0.125/2 = 0.0625 mole of CO2 will be produced.
30. (6 pts) Use words to rewrite the following chemical sentence stating what the equation represents in terms of individual molecules and in terms of moles of molecules.
B2O3 + CaF2(s) ---> BF3(s) + CaO(s)
Again, the first step is balancing.
B2O3 + 3CaF2(s) ---> 2BF3(s) + 3CaO(s)
In molecules, the sentence reads:
1 molecules of boron oxide react with 3 molecules of calcium
fluoride to produce 2 molecules of boron trifluoride and 3
molecules of calcium oxide.
In moles, the sentence reads:
1 moles of boron oxide react with 3 moles of calcium
fluoride to produce 2 moles of boron trifluoride and 3
moles of calcium oxide.