Download Chapter 9 Section 9.1: Team Learning Worksheet

Chapter 9
Section 9.1: Team Learning Worksheet
1. An individual coefficient does not tell us anything. What is important is the ratio between the
reactants and products. For example, suppose we were going to make cookies and a recipe told
us to use two eggs, some butter, some flour (etc.) and we would make some cookies. The fact
that we were told we needed two eggs is meaningless. It is helpful to know this number only if
we know the other amounts of ingredients needed (and the expected yield). This is an important
concept for the students to understand. Many students put too much importance in an individual
coefficient when doing stoichiometry calculations. For example, when given any mass of oxygen
and asked to determine the number of moles of hydrogen needed for a complete reaction, many
students will reply “2 moles of hydrogen” and cite the coefficient as their reason.
2. The “mole ratio” is the ratio of coefficients in a balanced chemical equation. For example, in the
balanced equation
N2(g) + 3H2(g) → 2NH3(g)
the mole ratio is 1 mole N2 to 3 moles H2 to 2 moles NH3. We use this ratio to calculate moles of
one substance from moles of a different substance. For example, if we have 4 moles of H2 and an
excess of N2, we can calculate the moles of NH3 produced as follows:
4 moles H2 ×
2 moles NH 3
= 2.67 mol NH3
3 moles H 2
3. 1.25 mol O2 is required.
1 mole NH3 ×
5 mol O 2
= 1.25 mol O2
4 mol NH 3
4. More water would be made than ammonia. Students need to determine the equations (and
balance them) and think about how to set up this problem. In addition, they are not given
amounts of reactants to use. The students can solve this in general (for example, starting with “x”
mol water), or they can use a number of their choosing. They should understand why they could
do this.
2H2 + O2 → 2H2O
30 mol H2 ×
2 mol H 2 O
= 30 mol H2O
2 mol H 2
N2 + 3H2 → 2NH3
30 mol H2 ×
2 mol NH 3
= 20 mol NH3
3 mol H 2
5. There are an infinite number of possible values for “c” and only one possible value for the
quantity “c/d.” This is because an equation can be balanced with any coefficients as long as the
numbers of each type of atom are the same on each side of the equation. However, this means
that the ratio between the coefficients is the same. For example, each of the following equations
are balanced properly.
2H2 + O2 → 2H2O
4H2 + 2O2 → 4H2O
H2 + ½O2 → H2O
8H2 + 4O2 → 8H2O
Notice that each has a different coefficient for H2O, for example. However, the quotient of the
coefficients between H2O and H2 is always 1.
Some students are likely to claim that there is only one value for “c” because they believe that
only an equation balanced in standard form (smallest whole numbers) is correct. If this is the
case, make sure the students understand why there can be any value for “c” before answering the
second question. Many students will find it difficult, but a discussion of this idea will help them
to understand the meaning of a mole ratio and why it is the ratio (and not the individual
coefficient) that is important.
Section 9.2: Team Learning Worksheet
1. a. 0.0575 mol MnO2; b. 0.230 mol HCl; c. 8.39 g HCl; d. 2.07 g H2O, 7.24 g MnCl2, 4.08 g Cl2;
e. 13.39 g = 13.39 g (8.39 + 5.00 = 2.07 + 7.24 + 4.08); f. yes; g. no
a. 500 g MnO2 ×
1 mol MnO 2
= 0.0575 mol MnO2
86.94 MnO 2
b. 0.0575 mol MnO2 ×
c. 0.230 mol HCl ×
4 mol HCl
= 0.230 mol HCl
1 mol MnO 2
36.458g HCl
= 8.39 g HCl
1 mol HCl
d. 0.0575 mol MnO2 ×
2 mol H 2 O
18.016g H 2 O
×
= 2.07 g H2O
1 mol MnO 2
1 mol H 2 O
0.0575 mol MnO2 ×
1 mol MnCl 2
125.846g MnCl 2
×
= 7.24 g MnCl2
1 mol MnO 2
1 mol MnCl 2
0.0575 mol MnO2 ×
1 mol Cl 2
70.90g Cl 2
×
= 4.08 g Cl2
1 mol MnO 2
1 mol Cl 2
2. 81.1 g O2 is required to react with 11.5 g of C6H14.
2C6H14 + 19O2 → 12CO2 + 14H2O
1 mol C 6 H 14
19 mol O 2
32.00g O 2
×
×
= 81.1 g O2
86.172g C 6 H 14
2 mol C 6 H14
1 mol O 2
11.5 g C6H14 ×
3. 41.2 g NaCl can be produced from 25.0 g of Cl2 and excess Na.
2Na + Cl2 → 2NaCl
25.0 g Cl2 ×
1 mol Cl 2
2 mol NaCl
58.44g NaCl
= 41.2 g NaCl
×
×
70.90g Cl 2
1 mol Cl 2
1 mol NaCl
4. 54.6 g carbon dioxide is produced. The students will have to determine and balance the
equations, as well as think about how to set up the problem.
2H2 + O2 → 2H2O
10.0 g H2 ×
1 mol H 2
1 mol O 2
×
= 2.48 mol O2
2.016g H 2
2 mol H 2
CH4 + 2O2 → CO2 + 2H2O
2.48 mol O2 ×
1 mol CO 2
44.01g CO 2
×
= 54.6 g CO2
2 mol O 2
1 mol CO 2
Section 9.3: Team Learning Worksheet
We have included choices for many of these questions to test the students’ understanding. We have
used these choices with students and found that the alternate choices are tempting to the students and
reveal misconceptions.
1. The answer is “e” (each would produce the same amount of product). Many students try to
answer this question without performing calculations. Choice “b” is chosen because this
combination has the greatest total number of moles of reactants; choice “c” is tempting because
the amounts of reactants are the same as the coefficients in the balanced equation. This problem
will not only help students to better understand the concept of limiting reactants, but also show
the students that they need to think about the problem and do some calculating before answering
the question.
a. H2 is limiting; 2 mol H2 ×
2 mol H 2 O
= 2 mol H2O
2 mol H 2
b. H2 is limiting; 2 mol H2 ×
2 mol H 2 O
= 2 mol H2O
2 mol H 2
c. Neither is limiting; 2 mol H2O produced
d. O2 is limiting; 1 mol O2 ×
2 mol H 2 O
= 2 mol H2O
1 mol H 2
2. The answer is “e” (none of the above). Students often memorize shortcuts concerning limiting
reactants (see the discussion to question 1 above). This problem shows the students that there is
no way to determine the limiting reactant without calculating. Having the students present
counterexamples to each of the choices is a good way to test their understanding.
3. The answer is “b” (A is the limiting reactant because you need 6 moles of A and have 4). This
problem confronts the misconception that the limiting reactant is the one that is present as the
least number of moles.
2 mol B ×
3 mol A
= 6 mol A
1 mol B
4. 22.5 g of water can be produced. The law of conservation of mass still holds true, but 40.0 g of
water is not produced because not all of the hydrogen reacts (oxygen is the limiting reactant).
2H2 + O2 → 2H2O
20.0 g H2 ×
1 mol H 2
2 mol H 2 O
18.016g H 2 O
×
×
= 179 g H2O
2.016g H 2
2 mol H 2
1 mol H 2 O
20.0 g O2 ×
1 mol O 2
2 mol H 2 O
18.016g H 2 O
×
×
= 22.5 g H2O
32.00g O 2
1 mol O 2
1 mol H 2 O
5. The maximum amount of C that can be produced is 2.0 moles. Many students will say 6.0 moles
(they will add 2.0 and 4.0). The students need to know that the amount of product is determined
by the least amount of product that can be formed, and they should understand why this is true.
6. The answer is “d.” If the molar mass of A is less than the molar mass of B (and we have equal
masses of each), there will be fewer moles of B than A. According to the balanced equation,
more B than A is required (by a 5:1 ratio), so B will be limiting. The key to the solution is the
word “must” in the choices. Many students will think that we need to know the molar masses of
A and B to answer this.