Download 258-261

Page 1 of 4
(continued)
for converting moles of H2 to moles of NH3:
2 mol NH3
1.30 E
mol H2 0.867 mol NH3
3E
mol H2
Check: Notice that the number of moles of NH3 that can be produced is
less than 1.30. This answer makes sense because it takes 3 moles of H2 to
yield 2 moles of NH3. Thus the number of moles of NH3 produced (0.867)
should be smaller than the number of moles of H2 consumed (1.30).
Focus Questions
Sections 9.1–9.2
1. What part of a chemical equation allows us to determine how much
product we can get from a known amount of reactants?
2. Why is it important to balance a chemical equation before answering any
questions about it?
3. Give at least three different ways to interpret the coefficient in a
chemical equation.
4. How does a mole ratio help to determine the amount of product produced
from a known number of moles of reactant?
5. What number of moles of O2(g) is required to react with 3.6 mol SO2(g)
in the following reaction?
2SO2(g) O2(g) n 2SO3(g)
Mass Calculations
Objective: To learn to relate masses of reactants and products
in a chemical reaction.
I
n the last section we saw how to use the balanced equation for a reaction
to calculate the numbers of moles of reactants and products for a particular case. However, moles represent numbers of molecules, and we cannot
count molecules directly. In chemistry we count by weighing. Therefore, in
this section we will review the procedures for converting between moles and
masses and will see how these procedures are applied to chemical calculations.
To develop these procedures we will consider the combustion of propane,
C3H8. Propane, when used as a fuel, reacts with oxygen to produce carbon
dioxide and water according to the following unbalanced equation:
C3H8(g) O2(g) n CO2(g) H2O(g)
PROBLEM SOLVING
Always balance the equation for
the reaction first.
258
What mass of oxygen will be required to react exactly with 44.1 g of propane?
To deal with the amounts of reactants and products, we first need the
balanced equation for this reaction:
C3H8(g) 5O2(g) n 3CO2(g) 4H2O(g)
Chapter 9 Chemical Quantities
Page 2 of 4
Next, let’s summarize what we know and what we want to find.
W H AT I F ?
Your lab partner has made the
observation that we always
measure the mass of chemicals
but then use mole ratios to
balance equations.
What We Know:
●
The balanced equation for the reaction
●
The mass of propane available (44.1 g)
What We Want to Calculate:
What if your lab partner decided
to balance equations by using
masses as coefficients? Is this
possible? Why or why not?
●
The mass of oxygen (O2) required to react exactly with all the propane
Our problem, in schematic form, is
96.1 g propane
? grams of O2
requires
Our overall plan of attack is as follows:
1. We are given the number of grams of propane, so we must convert to
moles of propane (C3H8), because the balanced equation deals in moles
rather than grams.
2. Next, we can use the coefficients in the balanced equation to determine
the moles of oxygen (O2) required.
3. Finally, we will use the molar mass of O2 to calculate grams of oxygen.
We can sketch this strategy as follows:
C3H8(g)
5O2(g)
44.1 g
C3H8
? grams of
O2
1
3
? moles of
C3H8
1
2
n
3CO2(g)
4H2O(g)
? moles of
O2
Thus the first question we must answer is, How many moles of propane are present in 44.1 g of propane? The molar mass of propane is 44.09 g (3 12.01 8 1.008). The moles of propane present can be calculated as follows:
1 mol C3H8
44.1 g C3H8 1.00 mol C3H8
44.09 g C3H8
2
Next we recognize that each mole of propane reacts with 5 mol of oxygen. This gives us the equivalence statement
1 mol C3H8 5 mol O2
from which we construct the mole ratio
5 mol O2
1 mol C3H8
that we need to convert from moles of propane molecules to moles of oxygen molecules.
5 mol O2
1.00 mol C3H8 5.00 mol O2
1 mol C3H8
9.3 Mass Calculations
259
Page 3 of 4
3
Notice that the mole ratio is set up so that the moles of C3H8 cancel and the
resulting units are moles of O2.
Because the original question asked for the mass of oxygen needed to react with 44.1 g of propane, we must convert the 5.00 mol of O2 to grams,
using the molar mass of O2 (32.00 2 16.00).
32.0 g O2
5.00 mol O2 160. g O2
1 mol O2
Therefore, 160 g of oxygen is required to burn 44.1 g of propane. We can
summarize this problem by writing out a “conversion string” that shows how
the problem was done.
1
3
2
5 mol O2
1 mol C3H8
32.0 g O2
160. g O2
44.1 g C3H8 1 mol C3H8
44.09 g C3H8
1 mol O2
PROBLEM SOLVING
Use units as a check to see that
you have used the correct conversion factors (mole ratios).
This is a convenient way to make sure the final units are correct. The procedure we have followed is summarized below.
C3H8(g)
5O2(g)
44.1 g
C3H8
160. g
O2
1
3
Use molar mass of
C3H8 (44.09 g)
Use molar mass of
O2 (32.0 g)
1
3
1.00 mol
C3H8
2
Use mole ratio:
5 mol O2
1 mol C3H8
2
n
3CO2(g)
4H2O(g)
5.00 mol
O2
This example illustrates in detail how to deal with masses of reactants in
chemical equations. To see whether these ideas are clear to you, try the following Self-Check Exercises.
Self-Check Exercise 9.2
What mass of carbon dioxide is produced when 44.1 g of propane reacts
with sufficient oxygen?
Self-Check Exercise 9.3
Calculate the mass of water formed by the complete reaction of 44.1 g of
propane with oxygen.
To illustrate how to deal with masses in a chemical equation, consider
the following example.
260
Chapter 9 Chemical Quantities
Page 4 of 4
Example 9.5
Mass Relationships in Chemical Reactions
Consider the reaction of powdered aluminum metal and finely ground iodine to produce aluminum iodide. The balanced equation for this vigorous chemical reaction is
2Al(s) 3I2(s) n 2AlI3(s)
Calculate the mass of I2(s) needed to just react with 35.0 g Al(s).
Solution
Let’s summarize what we know and what we need to calculate.
What We Know:
●
The balanced equation for the reaction
●
The mass of Al(s) available
What We Want to Calculate:
●
The mass of iodine (I2) required to react with all of the aluminum
(35.0 g)
Let’s summarize our plan:
Aluminum (left) and iodine
(right) shown at the top, react
vigorously to form aluminum
iodide. The purple cloud results from excess iodine
vaporized by the heat of the
reaction.
1. We first need to convert from 35.0 g aluminum to moles of aluminum because the balanced equation deals in moles rather than
grams.
2. We will use the balanced equation to find the moles of I2 required
to react with all of the aluminum.
3. Once we know the moles of I2(s) required, we will use the molar
mass for iodine to calculate the mass of I2(s) required.
A schematic for this strategy is
2Al(s)
3I2(s)
35.0 g
Al
? grams
I2
? moles of
Al
? moles of
I2
n
2AlI3(s)
Using the molar mass of aluminum (26.98 g), we will first calculate the
moles of aluminum present in 35.0 g Al:
1 mol Al
35.0 g Al 1.30 mol Al
26.98 g Al
From the balanced equation for the reaction we see that
2 Al requires 3 I2
(continued)
9.3 Mass Calculations
261