Experiment 6

Language of Chemistry (Part II): Balancing Chemical Reactions & Mole-

Mass Relationships

Learning Goals & Objectives

Like many fields, the study of chemistry is simplified by classifying information into groups having similar characteristics. Today’s laboratory activity focuses on chemical reactions and introduces students to the quantitative relationships between reactants and products in chemical reactions, which is known as stoichiometry.

Upon completing the exercises that follow, students will learn to: 1) recognize the parts of a chemical reaction; 2) identify different reaction classes from reactant and product configurations; 3) balance simple chemical reactions either by inspection or using the table method; 4) predict the identity of products formed in chemical reactions and 5) calculate quantities of reactants consumed or products formed in chemical reactions.

Introduction

Our working definition of chemistry has three core objectives, namely to investigate: 1) the structure of matter; 2) the composition of matter; and 3) the transformations (changes) that matter undergo. It is the changes that matter undergo that form the basis of chemical reactions.

In the chemistry language, chemical reactions function just like sentences do in your English courses. A sentence contains a sequence of words linked together to communicate a thought, action, or idea. In chemistry, reactants combine with one another under a set of conditions and are transformed into products. A properly written, balanced chemical reaction conveys a great deal of information in a concise manner. Let us review how “chemical sentences” are put together.

Chemical Reactions: The Basics

Despite the many reactions one typically encounters in a general chemistry course, the general form of any reaction is the same. We begin by looking at the parts of a chemical reaction along with additional information one is likely to encounter.

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Let us begin with a generic chemical reaction like the one shown below.

R yields→

P

The left side of any chemical reaction is where the reactants (R) or starting materials are found. The arrow is often associated with phrases like “yields”, “produces”, “forms”, “equals”, etc. Products (P) are found on the right side of a chemical reaction; these are the final substances formed from the starting materials or reactants.

Additional information about a chemical reaction (e.g., physical states, temperature, pressure, reaction time) is often included so that other scientists can achieve the same outcome. Physical states are symbolized as follows: solid (s), liquid (l), gas (g), and aqueous (aq) which means dissolved in water. Temperatures are typically given in Celsius (oC); pressure is given in units called atmospheres (atm); and time is given in minutes, hours, or days.

With the fundamentals in mind, the next step is to balance a chemical reaction. Why is this necessary? The answer can be traced back to the law of conservation of matter which states that matter can neither be created nor destroyed. For this law to be valid, one must know the amount of starting materials used in a reaction as well as understand how the reactants combine with one another to form products. All chemical reactions must be balanced; a reaction that is not balanced is of no use to any scientist.

How does one balance a chemical reaction? In the lecture course, you are currently using the table method to balance a chemical reaction. Recall that one uses coefficients to balance a chemical reaction; they are always found in front of a reactant or product formula. Consider the following general example between reactants A2 and B2 form the product A2B.

A2 + B 2 yields→

A 2 B

The table method tracks the individual number of atoms found on both sides of a chemical reaction. When a reaction is balanced, all the component atoms on both sides must be equal.

Our initial table takes the form shown below.

A2 + B 2 yields→

A 2 B

2 A 22 B 1

At this point the number of atoms of A is balanced while B is not balanced. We first introduce a coefficient of 2 to the right side of the reaction. Our reaction and table now have the configuration shown.

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A2 + B 2 yields→

2 A 2B

2 A 2 42 B 1 2

Notice how the introduction of the coefficient to the product side now changes both A and B on the right hand side of the reaction. Since B is now balanced on both sides of the reaction, we do not alter the coefficient of one (1) for B2 on the left side of the reaction. Instead, we introduce the coefficient of 2 in front of reactant A2 to balance the reaction system. Our final equation and table are presented below.

2 A 2 + B2 yields→

2 A2B

4 2 A 2 42 2 B 1 2

Notice that the coefficient for B2 appears to be missing in this first example; however, the coefficient for B2 is actually one (1). By convention, a coefficient of one (1) is routinely not included in most chemical reactions. So it is important to remember that a reactant or product in a balanced chemical reaction that has a missing coefficient is usually assumed to be one (1).

A common misconception on the part of some students involves attempting to balance a chemical reaction by either adding or deleting subscripts from the formula of a reactant or product. Do not alter the formulas of reactants or products since this changes the entire reaction into something completely different. We balance reactions using coefficients not subscripts.

Classifying Chemical Reactions

In the exercises that follow, we will examine six common reaction classes. As you read about each type of reaction, identify relationships between reactants and products. Write out the characteristics that allow one to identify each reaction class. By learning to recognize the reactant and product configurations, one readily identifies the reaction class.

1. Synthesis or Combination Reaction: A synthesis or combination reaction is one in which simple reactant substances (elements or compounds) combine to form a more complex single product. The general form of a synthesis reaction is shown below.

A + B yields→

AB

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Many ionic salts are prepared by reacting component elements together. We have already observed how table salt is produced; sodium metal reacts with chlorine gas to produce solid sodium chloride. The balanced equation for the process is shown below.

2 Na ( s ) + Cl 2 (g) yields→

2NaCl (s )

A second example of a combination reaction involves the production of carbonic acid, one of the main ingredients in soda. Water and carbon dioxide combine to form the product as depicted below.

H 2O (l ) + CO 2 (g) yields→

H 2 CO3 ( aq )

2. Decomposition Reaction: A decomposition reaction is easy to recognize if one looks for two things. The first way to recognize this reaction is to remember that decomposition is the opposite of synthesis. This statement means that a single complex reactant substance reacts to produce two or more simple product substances. How can a single substance react? That is a good question. Since compounds are stable substances, decomposition reactions typically require heat or electricity to get the process started; this is the second characteristic one looks for to identify this reaction class. In chemical reactions, heat is typically symbolized by the Greek letter delta (Δ). The general form of a decomposition reaction is as follows.

AB ∆→

A+B

Several general examples of decomposition reactions include the breakdown of mercury (II) oxide and calcium sulfite. The balanced chemical reactions for both substances are as follows. Notice in both examples that one starts with one reactant and heat is required to initiate the reaction.

2 HgO (s ) ∆→

2 Hg (l )+O2(g)

CaSO3 ( s ) ∆→CaO (s )+SO2(g)

3. Single Replacement or Single Displacement Reaction: In a single-replacement reaction, one element replaces (or displaces) another element in a compound. As illustrated by the general reaction,

A+BC → AC+B

when metallic zinc is placed into a solution of copper (II) chloride, the zinc replaces the copper.

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Zn ( s )+CuCl2 (aq) →ZnCl2 (aq) +Cu(s )

Another example of a displacement reaction involves reacting magnesium metal with hydrochloric acid to produce hydrogen gas and magnesium chloride.

Mg ( s )+2 HCl (aq) →MgCl 2 (aq) +H 2(g)

An important clue that indicates this class of reaction is that one of the starting reactants is an element while the other reactant is a compound.

4. Double Replacement or Double Displacement Reaction: The next general reaction class is known by several names, including double replacement, double displacement, or metathesis reaction. In a double replacement reaction, two elements or groups of elements in two different compounds exchange partners to form two new compounds. The general form of this reaction is as follows.

AB+CD → AD+C B

Metathesis reactions are seen most often in chemistry, often under different circumstances. For example, when two aqueous solutions are reacted one may produce an insoluble solid product. This type of reaction is called a precipitation reaction, named after the type of product formed. Some examples of precipitation reactions are illustrated below. In each case, notice how the two cation species switch partners to form new products.

BaCl 2 (aq )+Na2S (aq )→ BaS (s) + 2 NaCl (aq)

AlCl3 (aq )+3NH 4OH (aq)→ Al(OH)3 (s) + 3 NH 4 Cl (aq)

A second sub-group of chemical reaction that is often, but not always, seen in double displacement form is the acid-base neutralization reaction. Since you will spend a great deal of time with acid-base chemistry, let us examine this reaction class a little more closely to see how it works and how to recognize it.

5. Acid-Base Neutralization: As the name implies, acid-base reactions bring together two components, one an acid and the other a base. Recall that an acid is a substance that can donate

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a proton (H+) while a base is a compound or ion that can accept a proton. The products common to an acid-base reaction are a salt and often (but not always) water. Notice the variety of reactants and products that comply with our definition of acid (proton donor) or base (proton acceptor). Can acid-base reactions also belong to other reaction classes? Yes, indeed they can. We find that acid-base reactions often can be classified as a sub-group of double-replacement (4) and synthesis (1) reactions. The general form of a neutralization reaction, along with several examples, is shown below.

HA + BOH →H 2 O+BA

HNO 3(aq)+NaOH (aq)→ NaNO 3 (aq) + H 2O (l)

2 HCl (aq )+Na2CO3(aq)→ 2 NaCl (aq) + CO 2 (g) + H 2O (l)

HCl (aq )+NH 3 (g )→ NH 4 + Cl - (aq)

6. Oxidation-Reduction (Redox) Reactions: Another broad class of reaction that involves the transfer of electrons from one atom, ion or molecule to another atom, ion or molecule is referred to as oxidation-reduction (or redox) reaction. Electron movement is tracked by a change in oxidation number of specific reactant atoms, changing to a different oxidation number in the products.

Unlike the other general reaction classes described in this activity, oxidation-reduction reactions are often difficult to balance by inspection or even by the table method; this is because this type of reaction requires a special set of rules to balance the reaction. These rules will be presented and discussed in a future experiment. Oxidation-reduction processes are an important class of reaction that is found in many areas of science and engineering including metal refining, photosynthesis, energy-conversion devices, respiration, and corrosion.

Mass-Mole Calculations & Chemical Reactions

There is no simple one-step memory method of solving mole-mass problems. The proportions method and the product factor method give correct answers, but do you understand the calculative process that makes them work? It is our goal first to help you develop an understanding and

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appreciation of how such problems are solved. Then you will be able to properly use the product factor method.

When you are confronted with a problem in everyday life, there are specific mental processes and tools that you organize and employ as you seek a solution. Why not do the same thing when solving a chemistry problem? First try to identify the nature of the problem. What does the problem ask you to do or determine? Most mole-mass problems involve answering one of the following questions.

Typical Problem Types

a. How many moles of a substance can be produced from a fixed number of moles of reactants (mole-to-mole type)?

b. How many grams of a substance can be produced from a fixed number of grams of reactants (mass-to-mass type)?

c. How many grams of a substance can be produced from a fixed number of moles of reactants (mole-to-mass type)?

d. The reverse of items a, b, and c can also be asked. Given a quantity of product, how much reactant is required to produce it?

Next, organize in your mind and on paper the calculative tools you will need. These types of problems can be broken down into an approach referred to as the handy-dandy five step method; the methodology is summarized as follows.

1. Write the balanced chemical equation in formula form. Place the given and unknown quantities (with units) above each item in the chemical equation.

2. Calculate the formula mass (i.e., molecular weight) for the given and unknown quantities.3. Convert the given quantity to moles. “Moles are the vehicle that takes us everywhere we want

to go in chemistry.”4. Using the balanced chemical equation from step 1, convert the starting number of moles into

the number of moles of the final quantity sought.5. Convert the final quantity from moles to the desired unit, if necessary.

Why do I have to use this approach?

Yes, this process may seem lengthy, but it is the only way to gain a good understanding of the mass-mole relationship in chemistry. At this point we need to examine a chemical reaction and explore the information it provides. Consider the following reaction.

2 Mg + O2 yields→

2 MgO (s )

Mol. Wt. 24.31 32.00 40.31

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The above reaction expressed in words represents numerous possibilities. Look at the balanced equation and be bold; repeat aloud what it says.

2 atoms Mg + 1 molecule O2yields

→2 molecules MgO

2 moles Mg + 1 mole O2yields

→2 moles MgO

A chemical equation can be multiplied or divided by any number; recall that the arrow in this context means “equals” or = as in a mathematical equation. Therefore, the following are all true.

2(24.31 mg) Mg + 1(32.00 mg) O2yields

→2 (40.31 mg) MgO

2(24.31 g) Mg + 1(32.00 g) O2yields

→2 (40.31 g) MgO

2(24.31 lbs) Mg + 1(32.00 lbs) O2yields

→2 (40.31 lbs) MgO

2(24.31 tons) Mg + 1(32.00 tons) O2yields

→2 (40.31 tons) MgO

In each case, the Law of Conservation of Mass is obeyed. Do not take our word for it; run the numbers yourself to verify this fact. The sum of the masses of reactants equals the mass of the product. Note that the sum of the moles of reactants does not equal the sum of the moles of product. Moles are not necessarily conserved in a chemical reaction. These equations illustrate the connection between stoichiometric coefficients, moles, and mass. Study them carefully until you understand the relationships between them.

Mass-to-mole and mole-to-mole problems can be solved in two ways. One approach involves breaking the calculation into three stages; these are simply steps 3-5 of the handy-dandy method. The second approach is to present the calculation in one step using the product factor method.

Let us solve an example problem using both approaches and then compare and contrast the results to show the relationship between the methods of calculation. The problem reads as follows: If 12.0 g of Na2O is reacted with excess H2O, how many grams of NaOH will be produced by the reaction: Na2O + H2O → 2 NaOH?

This example is a mass-to-mass problem. It is selected specifically because it is one of the most difficult types of stoichiometric problems type you will encounter. Let us solve this exercise together.

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Problem Statement

If 12.0 g of Na2O is reacted with excess H2O, how many grams of NaOH will be produced by the reaction Na2O + H2O → 2 NaOH?

Handy-Dandy Five Step Method

Step 1: Write out the balanced chemical reaction with the given quantity (12.0 g Na 2O) and unknown item (? g NaOH) shown above each item as illustrated below.

12.0 g ? g

Na2O + H2O → 2 NaOH

Step 2: Calculate the Molecular Weights (MWs) for the given and unknown items

MW (Na2O ) = [2(23.00 g/mol) + 1(16.00 g/mol)] = 62.00 g/molMW (NaOH) = [1(23.00 g/mol) + 1(16.00 g/mol)+ 1(1.01 g/mol)] = 40.01 g/mol

At this point, we re-write steps 1 and 2 together to show how you will normally set-up the solution to this problem.

12.0 g ? g

Step 1: Na2O + H2O → 2 NaOH

Step 2: 62.00 g/mol 40.01 g/mol

MW (Na2O ) = [2(23.00 g/mol) + 1(16.00 g/mol)] = 62.00 g/mol

MW (NaOH) = [1(23.00 g/mol) + 1(16.00 g/mol)+ 1(1.01 g/mol)] = 40.01 g/mol

Step 3: Convert given quantity to moles because “Moles are the vehicle that carry us everywhere we want to go in chemistry.”

12.0 gNa2 O×( 1mol Na2O62.00g Na2O )=0.1935mol Na2O

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Step 4: Convert given number of moles from step 3 to the number of moles of the final quantity requested. Note: The balanced chemical reaction shows that 2 moles of NaOH are produced for every 1 mole of Na2O that react.

0.1935mol Na2O×(2mol NaOH1mol Na2O )=0.3870mol NaOH

Step 5: Convert the final quantity from moles to the unit desired.

0.3870mol NaOH ×( 40.01g NaOH1mol NaOH )=15.5g NaOH

Product-Factor Method

Begin the string of terms by writing what is given multiplied by the various factors and equating this expression to the desired quantity. Always include the units with each quantity.

given x factors = desired

On inspection of the general equation above, it is clear that the first factor must have the units g Na 2O in the denominator to cancel out the given unit, and that the final conversion factor must require the units of g NaOH in the numerator to match what is desired in the final answer.

12.0 gNa2 O×( ❑gNa2 O )×( g NaOH

❑ )=¿ NaOH

So our calculation begins to take shape and that is good, but you need to introduce factors that have mole units associated with them. Think about how you can expand these first two factors to include the mole terms. Hint: Does a mole represent a specific number of grams for a substance?

12.0 gNa2O×( 1mol Na2O62.00g Na2 O )×( 40.01 g NaOH

1mol NaOH )=¿ NaOH

What is wrong with the solution presented above? It is not correct. Can you see why?

Observe that your units do not cancel; here is yet another instance where units guide us in solving chemistry problems! You still need another factor that must use the mole : mole relationship between Na2O and NaOH such that all the units cancel out. What will the factor look like?

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If you came up with the factor (mol NaOHmol Na2O ) as the result, you are correct. Notice that the units will

cancel only if the factor is presented in this fashion. The final calculation is shown below.

12.0 gNa2 O×( 1mol Na2O62.00g Na2O )×( 40.01 g NaOH

1mol NaOH )×( 2mol NaOH1mol Na2O )=15.5g NaOH

Note that we obtain the same answer using both methods. Let’s rearrange the preceding equation. If we reverse the order of the last two factors as shown in the graphic below, suddenly we can see that the product-factor method is nothing more than steps 3-5 of the handy-dandy five step method performed on one line. The three different factors are identified here with their counterparts in the handy-dandy method.

12.0 gNa2 O( 1mol Na2 O62.00g Na2O ) ( 2mol NaOH

1molNa2O ) ( 40.01 gNaOH1mol NaOH ) ¿15.5g NaOH

Step 3 Step 4 Step 5

Which method should you use? While both approaches lead to the correct answer, in the beginning, it is best to use the handy-dandy method since it requires students to proceed through the process of understanding how stoichiometry works rather than relying on simply cancelling units. Once you feel comfortable with the concept, then use the product factor method since it can be direct and quick.

Experimental

Prior to laboratory, consult the internet or a general chemistry textbook to find examples or explanations of how to recognize and balance various types of chemical reactions. Review how to solve mole-mass problems.

Today you will study and balance by inspection five representative reaction types. This experiment does not consider oxidation-reduction reactions. They will be presented and discussed in another experiment.

Next, you will employ the beaker analogy to predict the products for several reactions. From the product configurations, you will state what reaction class is observed; and then balance each reaction using the table method. Finally, one will gain experience with mole-mass relationships in chemical reactions by calculating formula masses and performing stoichiometric calculations.

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This activity will require you to have your lecture textbook, laptop computer, calculator, and several extra sheets of paper to use for solving the problems presented.

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Experiment 6 Language of Chemistry (Part II)

Name: Section:

Instructor: Date:

Data Report Sheet

Part I: Examine each chemical reaction below. Compare and contrast the reactant and product configurations so that you can quickly classify reactions as to a specific type. Balance each reaction by inspection.

1. Synthesis or Combination Reactions:

A + B →A B

a. ___ Mg (s) + ___ O2 (g) → ___ MgO (s)

b. ___ Al (s) + ___ O2 (g) → ___ Al2O3 (s)

c. ___ N2O5 (g) + ___ H2O (l) → ___ HNO3 (aq)

d. ___ Na2O (s) + ___ H2O (l) → ___ NaOH (aq)

e. ___ Al (s) + ___ Br2 (l) → ___ AlBr3 (s)

2. Decomposition Reactions:

AB ∆→

A+B

f. ___ KClO3 (s) ∆→

___ KCl (s) + ___ O2 (g)

g. ___ H2O (l) ∆→

___ H2 (g) + ___ O2 (g)

h. ___ (NH4)2Cr2O7 (s) ∆→

___ Cr2O3 (s) + ___ N2 (g) + ___ H2O (l)

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Experiment 6 Language of Chemistry (Part II)

Name: Section:

Instructor: Date:

Data Report Sheet

3. Single-Replacement Reactions:

A+BC → AC+B

i. ___ Fe (s) + ___ H2SO4 (l) → ___ Fe2(SO4)3 (aq) + ___ H2 (g)

j. ___ Zn (s) + ___ HCl (l) → ___ ZnCl2 (aq) + ___ H2 (g)

k. ___ Fe (s) + ___ H2O (l) → ___ Fe2O3 (s) + ___ H2 (g)

l. ___ Cl2 (g) + ___ NaBr (aq) → ___ NaCl (aq) + ___ Br2 (l)

4. Metathesis or Double-Replacement Reactions:

AB+CD → AD+C B

m.

___ FeCl3 (s) + ___ NH4OH (aq) → ___ Fe(OH)3 (l) + ___ NH4Cl (aq)

n. ___ CaO (s) + ___ HCl (aq) → ___ CaCl2 (aq) + ___ H2O (l)

o. ___ BaCl2 (aq) + ___ AgNO3 (aq) → ___ Ba(NO3)2 (aq) + ___ AgCl (s)

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Experiment 6 Language of Chemistry (Part II)

Name: Section:

Instructor: Date:

Data Report Sheet

5. Acid-Base Neutralization Reactions:

Acid+Base →Salt+Water

p. ___ H2SO4 (aq) + ___ Ba(OH)2 (aq) → ___ BaSO4 (aq) + ___ H2O (l)

q. ___ HBr (aq) + ___ Al(OH)3 (aq) → ___ AlBr3 (aq) + ___ H2O (l)

Part II: Predict the products that form when the following reagents interact in water, and then balance each reaction by inspection. What class of chemical reaction does each represent?

Sample Problem

PbCl2 + NaI yields→

Strategy (The Beaker Analogy): Here are the steps one takes to solve this kind of problem.

1. We deconstruct (take apart) the starting substances (called the reactants).2. Draw a picture showing the ions that make up each reactant in a beaker of water.3. Now we determine what possible new substances (known as candidates) will form

from the reactants. Hint: Recall that ions in solution are free to move; reacting with one another only if their charges are opposite (i.e., n + m - and not n + m + or n - m - ).

4. Evaluate what candidates lead to new substances. 5. Write out the chemical formulas for the final substances (termed the products), and

then balance the final reaction. 6. Identify the reaction classification.

Solution

Example Reactants Products Type of Reaction (Classification)PbCl2 + 2 NaI yields

→PbI2 + 2 NaCl Metathesis or Double-Replacement

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Experiment 6 Language of Chemistry (Part II)

Name: Section:

Instructor: Date:

Data Report Sheet

Part II: Predict the products that might form when the following reagents interact in water, and then balance each reaction using the table method. What class of chemical reaction does each represent?

6. Extra for Experts: Predicting Products & Classifying Reactions

Item Reactants Products Type of Reaction (Classification)

r. ___Fe + ___ CuSO4 yields→

s. ___(NH4)2SO4 + ___ BaCl2 yields→

t. ___Ca(OH)2 + ___ H3PO4 yields→

u. ___SO2 + ___ H2O yields→

v. ___ZnCrO4 + ___ AgNO3 yields→

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Experiment 6 Language of Chemistry (Part II)

Name: Section:

Instructor: Date:

Data Report Sheet

Use this page to show your work in balancing reactions (items r through v) using the table method.

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Experiment 6 Language of Chemistry (Part II)

Name: Section:

Instructor: Date:

Data Report Sheet

Part III: Calculate the quantity represented in each of the following questions. You may assume that in each scenario, the reaction goes 100% to completion; this means that reactants are transformed into products.

i. In reaction b, how many grams of product theoretically can be produced from 48.0 grams molecular oxygen?

ii. In reaction f, how many moles of KClO3 are required to produce 33.0 moles of molecular oxygen?

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Experiment 6 Language of Chemistry (Part II)

Name: Section:

Instructor: Date:

Data Report Sheet

Part III: Calculate the quantity represented in each of the following questions. You may assume that in each scenario, the reaction goes 100% to completion; this means that reactants are transformed into products.

iii. In reaction i, how many grams of H2 and Fe2(SO4)3 can be produced from 75.0 grams of iron and excess H2SO4?

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