CHM 101 GENERAL CHEMISTRY

FALL QUARTER 2008

Section 2

Lecture Notes – 10/20/2008

(last revised: 10/19/08, 10:30 PM)

Review: We have covered the following topics in Chapter 3:

o Counting by Weighing

o Calculating a Weighted Average

o What is a Mole?

o Converting Moles to Mass and Mass to Moles

o Calculating % Composition from a Chemical Formula

o Determining a Chemical Formula from the % Composition

o Writing a Chemical Equation

o Balancing a Chemical Equation

Now we’ll turn to Mass to Moles and Moles to Mass Conversions among the reactants and products in a balanced chemical equation. The prototypical problem is that you are given the mass of one component (Call it A.) and asked to calculate the mass of another component (Call it B.). The way it is solved in the book is:

o Balance the equation for the reaction (if it is not already balanced).

o Determine the number of moles of A from its mass.

o Determine the number of moles of B by multiplying the number of moles of A by the appropriate mole ratio determined from the balanced equation.

o Determine the mass of B from the number of moles of B.

While the method in the text is perfectly valid, it could be a bit confusing. Recall that we deal with amounts of reagents that are counted as moles but measured in grams. I will show you how to write an equation for the number of moles of B in terms of the number of moles of A, and use it to solve the problem. You will see that we do the very same arithmetic as is demonstrated in the text:

o Balance the equation for the reaction (if it is not already balanced).

o Write down an equation that gives the number of moles of B in terms of the number of moles of A. (i. e., Use the appropriate mole ratios.)

o Plug the mass of A and the molar mass of A into this equation and solve for the number of moles of B.

o Determine the mass of B from the number of moles of B and the molar mass of B.

3.9 Stoichiometric Calculations: Amounts of Reactants and Products

• Now we extend the concept of counting by weighing to chemical reactions.
• The first example in the text starts with the question: What mass of oxygen (O₂) will react with 96.1 grams of propane (C₃H₈) (when propane burns in the presence of oxygen)?

o The first step is to write the balanced equation:

C₃H₈(g) + 5O₂(g) = 3CO₂(g) + 4H₂O(g)

o The balanced equation gives us the following relationships among the numbers of moles of its products and reactants (where the N’s represent numbers of moles):

o Then the number of moles of O₂ is:

o If we use the symbol, m, to represent the mass of a component in this reaction and the symbol, MM, to represent a molar mass, we can write:

o Then the mass of O₂ is:

o The result is that it takes 349. g of O₂ to burn 96.1 g of C₃H₈.

• The text extends this example by asking: What mass of carbon dioxide (CO₂) is produced when 96.1 grams of propane (C₃H₈) burns in the presence of oxygen (O₂)?

o The balanced equation for the reaction gives us:

o And we can write:

o Then the mass of CO₂ produced is:

• Problems to work out on the whiteboard:

o Sample Exercise 3.16 (p. 105): What mass of CO₂ can be absorbed from the living environment of a space capsule by 1.00 kg of LiOH?

o Sample Exercise 3.17 (pp. 105-6): Baking soda (NaHCO₃) can be used as an antacid to neutralize excess hydrochloric acid (HCl) secreted by the stomach:

NaHCO₃(s) + HCl(aq) = NaCl(aq) + H₂O(l) + CO₂(aq)

Similarly, milk of magnesia (Mg(OH)₂) can also be used to neutralize excess HCl:

Mg(OH)₂(s) + 2HCl(aq) = MgCl₂(aq) + 2H₂O(l)

Gram for gram, which is the more effective antacid, baking soda (NaHCO₃) or milk of magnesia (Mg(OH)₂)?

3.10 Calculations Involving a Limiting Reactant

• In Section 3.9 we calculated stoichiometric quantities of reagents involved in a given chemical reaction. For example, we found that if we had 96.1 g of propane, we would require 349. g of oxygen to burn it completely with no leftover oxygen (or propane). In other words, in the combustion reaction of propane with oxygen, the 96.1 g of propane and the 349. g of oxygen are stoichiometric quantities, and the mixture is a stoichiometric mixture.
• Suppose instead, we mixed the 96.1 g of propane with 500. g of oxygen and allowed the reaction to run. 1) Would all the propane react? 2) Would all the oxygen react? 3) What, if anything, would be left over? The short answers are:
1. Yes. We started with plenty of oxygen to react with all 96.1 g of the propane.
2. No. After 349. g of oxygen reacts, there is nothing left to burn.
3. There would be 151. g of oxygen left over. All the propane would have been consumed.

We would identify propane as the limiting reactant because all of it gets used up in the reaction.

• If we had mixed 120.0 g of propane with 349. g of oxygen and allowed the reaction to run, we would instead get these answers:
1. No. Once 96.1 g of propane burns, there is no more oxygen for the remaining propane.
2. Yes. We started with plenty of propane to react with all 349. g of the oxygen.
3. There would be 23.9 g of propane left over. All the oxygen would have been consumed.

This time the limiting reactant is oxygen.

• The text introduces limiting reactant calculations with a discussion of the reaction that produces hydrogen (H₂) as an intermediate reactant for the manufacture of ammonia (NH₃) by the Haber process:

CH₄(g) + H₂O(g) = 3H₂ + CO(g)

From the equation and the picture for the reaction, one can easily see that any mixture of equimolar or equimolecular quantities of CH₄ and H₂O is a stoiochiometric mixture. Three such mixtures are illustrated in Figure 3.9 in the text:

But we need to count our moles of CH₄ and H₂O by weighing. The text asks how many grams of water will be needed to react exactly with 2.50 x 10³ kg (2.50 x 10⁶ g) of CH₄. From the reaction, we can construct the relationship between the number of moles of H₂O and the number of moles of CH₄:

Then we can substitute the mass and the molar mass of CH₄ and compute the result for the number of moles of H₂O:

Thus the stoichiometric amount of H₂O is:

If we were to mix 2.50 x 10⁶ g (2.50 x 10³ kg) of CH₄ with 2.81 x 10⁶ g (2.81 x 10³ kg) of H₂O and allow them to react, we would run out of both reactants at the same time, and none of either would be left over. But if we used more water in the mixture (say 3.00 x 10⁶ g), we would use up all the methane, but there would be water left over. Figure 3.10 from the text illustrates a mixture of 6 molecules of CH₄ and 9 molecules of water, where water is also in excess,

and Figure 3.11 pictures the same collection of atoms after the reaction has run to completion. You will notice that the 3 excess water molecules are still there, but that the 6 methane molecules and the 6 waters that reacted with them are all gone, replaced by the products, hydrogen (H₂) and carbon monoxide (CO).

Clearly, methane is the limiting reagent in this situation. Notice that it not only limits the amount of methane that can react, but it also limits the amounts of product that can form. We will now see examples where we will need to determine which of two (or more) reactants is the limiting reactant.

• The next example in the text is the final step for the manufacture of ammonia (NH₃) by the Haber process:

N₂(g) + 3H₂(g) = 2NH₂(g)

We can write out the equations that relate the numbers of moles of reactants and products:

The example in the book mixes 5 molecules of N₂ with 9 molecules of H₂. Let’s instead consider 5 moles of N₂ mixed with 9 moles of H₂. With these quantities, which is the limiting reagent? Let’s solve for the number of moles of N₂ that will react with our 9 moles of H₂.

3 moles of N₂ will react with the entire 9 moles of H₂, and 2 will remain unreacted. Now let’s solve for the number of moles of H₂ needed to react with all 5 miles of N₂.

Clearly the 9 moles of H₂ are not sufficient to react with all the N₂.By either calculation, the H₂ will all be used up. Therefore H₂ is the limiting reagent in the case where 9 moles of H₂ are mixed with 5 moles of N₂, and in the case pictured in Figure 3.12 where 9 molecules of H₂ are mixed with 5 molecules of N₂, producing 6 molecules of NH₃, with 2 molecules of N₂ left over.

• Let’s take another look at the same reaction, but with new quantities of H₂ and N₂, this time at a manufacturing size scale. We mix 25.0 kg of H₂ with 5.00 kg of N₂, and ask, how much NH₃ should we expect to make? Again, the approach shown in the book is perfectly valid, but let’s do it another way. If N₂ is the limiting reagent, we will make 1 mole of NH₃ for each 2 moles of N₂:

On the other hand, if H₂ is the limiting reagent, we’ll make 1 mole of NH₃ for each 2/3 moles of H₂:

Since the amount of H₂ is more limiting than the amount of N₂, H₂ is the limiting reactant, and we will produce 1.653 x 10³ moles of NH₃. Then the mass of NH3 that we will make is given by:

• Let’s now look at Sample Exercise 3.18 (p. 110). We pass gaseous ammonia (NH₃) over solid copper (II) oxide (CuO) at elevated temperature to produce nitrogen (N₂) gas and the byproducts water (H₂O) vapor and solid copper (Cu). If we start with 18.1 g of ammonia and 90.4 g of copper (II) oxide, which is the limiting reactant and how much nitrogen will we make? We’ll start by writing and balancing the equation for the reaction:

2NH₃ (g) + 3CuO (s) = N₂ (g) + 3Cu (s) + 3 H₂O (g)

Next we’ll obtain the relationships among the numbers of moles of NH₃, CuO, and N₂:

If NH₃ is the limiting reagent, we can calculate the number of moles of N₂ to be:

On the other hand, if CuO is the limiting reagent, we will obtain:

Clearly, CuO is the limiting reagent, and the mass of N₂ we will obtain is:

• Percent Yield: An important variation on the limiting reactant problem is the problem of percent yield. In its simplest form, percent yield is calculated as:

Sometimes you will have a single reactant. Or you might have a stoichiometric mixture of reactants. Or you might be told which is the limiting reactant. In these cases you can calculate the amount of product you expected by the simple methods in Section 3.9. Other times you will need to identify the limiting reactant before you can calculate the expected amount of product. For example, if you ran the experiment described in Sample Exercise 3.18, but the actual amount of N₂ you obtained were 6.63 g, instead of the 10.6 g you calculated, your percent yield would be:

• Let’s try Sample Exercise 3.19 (pp. 111-2) from the text. Methanol (CH₃OH), otherwise known as methyl alcohol. It can be manufactured from gaseous carbon monoxide (CO) and hydrogen (H₂):

2H₂ (g) + CO (g) = CH₃OH (l)

Suppose 68.5 kg of CO (g) is reacted with 8.60 kg of H₂ (g). Calculate the theoretical yield of CH₃OH. (This is the limiting reactant problem.) If 3.57 x 10⁴ g of CH₃OH are actually produced, what is the % yield? We start with the balanced reaction and write the relationships among the moles of H₂, CO, and CH₃OH:

If H₂ is the limiting reactant, then the number of moles of CH₃OH that theoretically could be produced is:

However, if CO is the limiting reactant, then the number of moles of CH₃OH that theoretically could be produced is:

So H₂ is the limiting reactant, and the theoretical yield (in grams) of CH₃OH is:

Then the % Yield of CH₃OH is: