Module 3 Notes Name:Chemical Equations:

Are a way to represent chemical reactions on paper.

Equations must be balanced in order to satisfy the Law of Conservation of Mass:Atoms can be neither created nor destroyed in an ordinary chemical reaction, so there must be the same number of atoms on both sides of the equation.

Look at the Chemical Equation below:

2NaOH Na2O + H2OThese numbers are found in a chemical equation: Subscripts are the small numbers to the lower right of chemical symbols. Subscripts represent the number of atoms of each element in the molecule.

Coefficients are the large numbers in front of chemical formulas. Coefficients represent the number of molecules of the substance in the reaction.

In both cases when no number is represented the assumption is ONE.

So in the above equation there are 2 NaOH molecules, 1 Na2O molecule and 1 water molecule in order to balance the left and right sides. Both the reactant and product sides contain 2 sodium atoms, 2 oxygen atoms and 2 hydrogen atoms.

Balancing Equations By Inspection

***The order in which the following steps are performed is important.While shortcuts are possible, following these steps in order is the best way to be sure you are correct***

When balancing equations, you only have control over the Coefficients. Never change subscripts. 

1. Check for Diatomic Molecules - H2 - N2 - O2 - F2 - Cl2 - Br2 - I2 If these elements appear by themselves in an equation, they must be written with the subscript 2

2. Balance Metals3. Balance Nonmetals4. Balance Oxygen5. Balance Hydrogen 6. Recount All Atoms

If the atoms are not balanced at this point, there is a problem somewhere. Work your way back up the steps, from bottom to top, until you find the problem, and correct it.

7. If every coefficient will reduce, rewrite in the simplest whole-number ratio. An equation is not properly balanced if the coefficients are not written in their lowest whole-number ratio.  

Eg1. Balance the following equation: sodium hydroxide sodium oxide + water Step 1: change this word equation into a chemical equation using your periodic table and ion charge sheet

NaOH Na2OH + H2O1

Step 2: run through the check list above No diatomics Metals – inorder to balance Na2 on products I must put

coefficient 2 in front of NaOH in reactants2NaOH Na2OH + H2O

Non-metals – oxygens and hydrogens are balance on both sides Double check the number of each atoms on each side it looks

ok.Step 3 : state final answer

2NaOH Na2OH + H2OEg2. Balance: iron + oxygen iron (III) oxide Step 1:Fe + O2 Fe2O3 Step 2:-oxygen is a diatomic O2 unlike Fe

-There are two Fe atoms in products by subscript 2 so we must balance that in reactants by placing coefficient 2 in front of Fe

2Fe + O2 Fe2O3-It’s a two step process to balance the oxygens…since 2 and 3 aren’t multiples of each other, we multiply them in order to match the oxygens on each side. So place coefficient 3 on reactant side and coefficient 2 on product side

2Fe + 3O2 2Fe2O3This balances the oxygens at 6 on each side.- on the atom recount can you see that the Fe atoms no longer balance? So we need to make a final adjustment change the 2Fe on reactant side to 4Fe

4Fe + 3O2 2Fe2O3Step 3:4Fe + 3O2 2Fe2O3

Eg 3. Balance: carbon dioxide + water glucose + oxygen Step 1:CO2 + H2O C6H12O6 + O2Step 2:-oxygen is diatomic

-no metals-balance C on reactants by coefficient 66CO2 + H2O C6H12O6 + O2-balance hydrogen (I think you might see the problem if I balance Oxygen first) coefficient 6 for H2O6CO2 + 6H2O C6H12O6 + O2-now balance the oxygen. There are 18 oxygens in total on reactant side. Already 6 oxygens in glucose on product side, so in order to get the remaining 12 on products place coefficient 6 in front of O26CO2 + 6H2O C6H12O6 + 6O2-double check of each atom looks good

Step 3:6CO2 + 6H2O C6H12O6 + 6O2

Eg.4 iron (II) sulfide + hydrochloric acid iron (II) chloride + hydrogen sulfide Step 1:FeS + HCl FeCl2 + H2SStep 2:- no diatomics

- Fe so far is balanced- balance Cl FeS + 2HCl FeCl2 + H2S- sulfur is balanced- hydrogen is balanced

Step 3:FeS + 2HCl FeCl2 + H2SBalancing Equations using the Algebraic Method:

This is a useful alternative for those of you who have trouble just guessing and checking.

Also helps with those more complicated equations that need balancing

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However if your algebraic abilities are on the weak I would say keep tackling the trial and error method

Example 1 I2O5 + CO I2 + CO2

Steps:1. Write the skeleton Equation and represent coefficients as letters - a,

b, c etc.a I2 O5 + b CO c I2 + d CO2

2. Make algebraic equations in terms of atoms. Remember reactants equal products. There are as many equations as there are types of atomsAtoms Reactants ProductsI 2a 2cO 5a + b 2dC b d

3. Look over the equation and arbitrarily assign 1 to a coefficient (eg. a = 1)

Let a =1Note: could be b=1 or c=1

4. Solve the system of equations knowing reactants equals products2a = 2c 5a +b = 2d b = d2(1)=2c 5(1)+b =2d 5 = dc=1 5+b = 2b

b = 5

5. Insert coefficients into final equation1I2 O5 + 5 CO 1I2 + 5 CO2

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Types of Chemical Reactions:

A Chemical Reaction is a process in which substances change into other substances

You know a chemical reaction takes place if one or more of these occur: 1. Color changes - Different combinations of molecules reflect light

differently. A color change indicates a change in molecules. 2. Heat content changes - In all chemical reactions, the heat content of the

reactants and the content of the products is never the same. Sometimes the difference is great and can be easily detected; at other times the difference is slight and is more difficult to detect.

3. A gas is produced - Whenever a gaseous product forms in a liquid solution, bubbles can be seen. A colorless gas produced in a reaction of solids is much harder to detect.

4. A precipitate forms - Precipitates are insoluble products formed by a reaction taking place in a liquid solution. This insoluble product will eventually settle to the bottom, but might immediately appear by turning the clear solution cloudy.

There are Eight types of Chemical Reactions

1. Decomposition Reactions A compound breaks into parts.

Eg. 2H2O 2H2 + O2 compound element + element

2. Synthesis Reactions Elements or compounds are joined together.

Eg. 2H2 + O2 2H2O 6CO2 + 6H2O C6H12O6 + 6O2

element + element compound 3. Single Displacement Reactions

A single element replaces an element in a compound. Eg. Zn + 2HCl H2 + ZnCl2  

element + compound element + compound 4. Double Displacement Reactions

An element from each of two compounds switch places. Eg. H2SO4 + 2NaOH Na2SO4 + 2H2O

compound + compound compound + compound 5. Neutralization Reactions

Special types of double displacement reactions that involve the reaction between an acid and base to form a salt and water.

Heat is usually given off in neutralization reactions. Eg. A suspension of solid magnesium hydroxide in water is widely used

as an antacid to neutralize excess stomach acid: Mg(OH)2 (s) + 2HCl (aq) MgCl2 (aq) + 2H2O (s)

acid + base salt + water 6. Precipitation Reactions

Aqueous reactions that involve the formation of a precipitate (solid). Eg. 2KI (aq) + Pb(NO3)2 (aq) 2KNO3 (aq) + PbI2 (s)

soluble compound + soluble compound insoluble compound 7. Combustion Reactions

An organic compound (containing only carbon, hydrogen or oxygen) combines with oxygen. The products of combustion are always carbon dioxide and water.

Eg. CH4 + 2O2 CO2 + 2H2O hydrocarbon + oxygen carbon dioxide + water

8. Oxidation-Reduction Reactions4

Any reaction in which elements experience a change in oxidation number. Eg. S + O2 SO2 In this reaction, sulfur and oxygen both have an oxidation number (aka combining capacity) of zero before the reaction. After the reaction, sulfur is +4 and oxygen is -2.

Recognizing the 8 different reaction types you can start to predict products given reactants. I hope you can start to recognize the patterns. It is pretty straight forward for each reaction type except for the single replacement reactions. For these reaction types you can’t assume the reaction occurs. You have to consider the activity series. Activity Series

A listing of metals (and hydrogen) in order of decreasing activity

Metal Reactivity Metal Ion

Lithium Potassium Calcium Sodium Magnesium Aluminum Manganese Zinc Chromium Iron Lead HydrogenCopper Mercury Silver Platinum Gold

Non metal

FluorineChlorineBromineIodine

Most Reactive - - - - - - - - - - - - - - Least Reactive

Li+ K+ Ca2+ Na+ Mg2+ Al3+ Mn2+ Zn2+ Cr2+, Cr3+ Fe2+, Fe3+ Pb2+ H+Cu2+ Hg2+ Ag+ Pt2+ Au+, Au3+

F2Cl2Br2I2

Here’s the rule to use the activity series. A single or double replacement reaction will only occur if the REPLACING metal is ABOVE the existing metal in the compound.

Eg1. Determine the products CuCl2 + AlStep 1: Locate Cu and Al on the activity series – notice Al is above Cu so YES

the standard single replacement reaction occurs.Step 2: Write the complete equation – don’t forget to consider combining

capacities in new compounds formed: CuCl2 + Al Cu + AlCl3

Eg2. Determine the products CuCl2 + AuStep 1: Locate Cu and Au on the activity series – notice Au is below Cu so you

may thing the products are AuCl + Cu, but actually no reaction occursStep 2: Write the complete equation:

CuCl2 + Au no reaction

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Energy Changes in Chemical Reactions The heat effect - H (change in heat)

In chemical reactions bonds are broken or formed. Energy is needed (absorbed) to break bonds and energy is given off to form bonds.

For example in the decomposition reaction of HCl

HCl + energy H2 + Cl2 or HCl H2 + Cl2 - energy

Energy is needed by the reactants for the reaction to occur, that is, energy is absorbed. HCl has less energy than the separated H and Cl atoms. This is known as an endothermic reaction. (heat enters the reaction) This type of reaction feels cold to touch (the reactants take the heat away from you) H is positive (heat enters) when on the reactant side

The opposite is seen in a synthesis reaction such as Mg

Mg + O2 MgO2 + energy or Mg + O2 – energy MgO2

Energy is given off by the products when the reaction occurs. Mg and O have more energy than MgO2

This is known as an exothermic reaction. (heat exits the reaction) This type of reaction feels hot to touch (products are releasing heat) H is negative (heat exits) when on the reactant side

The energy gain or loss (H) is expressed in kilojoules (kJ) per mole and we can show the energy change using an energy diagram.*to determine if a reaction of endothermic or exothermic your first step should be to write the equation so that H is on the reactant side.

Eg 1. Given CH2 + 2 O2 CO2 + 2H2O + 891kJ a) Write the balanced equation 2 other ways (ie move H) and b) Draw the energy diagram and identify it as an exo or endo rxn

a) CH2 + 2O2 – 891 kJ CO2 + 2H2O orCH2 + 2O2 + 2H2O ∆H = -891kJThis is an exothermic reaction

b) The graph is only a rough representation of the energy change over timethe downwards arrow represents an energy loss – in other words an exothermic reaction

CH2 + 2O2

∆H = -891kJ Energy

CO2 + 2H2O

Reaction proceeds over time

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Eg2. Given KClO3 (s) + 41.4kJ K+ (aq) + ClO3-(aq)

a) Write an alternate form of the balanced equation and b) draw the energy diagram and identify it as an exo or endo rxn

a) KClO3(s) K+(aq) + ClO3(aq)

- - 41.4kJThis is an endothermic reaction

b) This graph represent an energy gain – an endothermic reaction

K+(aq) + ClO3(aq)

-

Energy ∆H = 41.4 kJ

KClO3(aq)

Reaction proceeds over time

ENTHALPY (H) = the heat contained in a system (products or reactants), so H is the change in heat or enthalpy. In other words H = Hproducts – Hreactants

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Stoichiometry

This is what we have been working towards since the beginning of the course. I have worked with you on 2 basic concepts, the mole and balanced equations. Now we are going to join them. In essence,

The mole + balanced equations = stoichiometry

Officially…

“Stoichiometry is the relationship between the amount of reactants used in a chemical reaction and the amount of products produced by the reaction”.Stoichiometry allows us to predict the amount of reactants given a certain amount of product or visa versa

First I’d like to revisit the idea of the coefficient in a balanced equation. What does it mean?

Eg. 2H2 + O2 2H2O

Remember the coefficient represents the number of molecules. So 2 molecules of hydrogen gas react with 1 molecule of oxygen to produce 2 molecules of water.

This equation is still balanced if I have a hundred times the number of molecules.

200H2 + 100O2 200H2O

Or I could have 6.02x1023 time the number of molecules

2(6.02x1023)H2 + 6.02x1023 O2 2(6.02x1023)H2O

Remember 6.02x1023 of any thing is 1 mole.

Therefore the coefficients can also represent the number of moles

2 mol H2 + 1 mol O2 2 mol H2O

Application:

Eg1. Consider N2 + H2 NH31. Balance this equation

Before starting any question now the first thing you have to do is balance the equation

N2 + 3H2 2NH3

2. If 15 molecules of H2 react, how many N2 molecules are needed?

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BEWARE with such a simple question you can probably see the relationship, or worse yet you may be tempted to do some kind of ratio calculation:

3H2 = 1N215H2 ?By cross multiplication, ? = 5 molecules N2

BUT DON’TGo back to your trusty Unit Conversions you will thank me later!

Step 1: From the balance equation pick out the relationship between H2 and N2 (as asked in question)

1 molecule N2 reacts with 3 molecules H2This information gives us 2 possible conversion factors

1 molecules N2 or 3 molecules H23 molecules H2 1 molecules N2

And you thought we were done with conversion factors!

Step 2: Create your multi-step calculation, starting with what we are giving and choosing the conversion factor that cancels out the “right” molecules

15 molecules H2 x 1 molecule N2 = 5 N2 molecules are needed3 molecules H2

**notice molecules of H2 cancel out**3. How many moles of NH3 are produced when 18 moles of H2 are reacted?

We are still using the same balanced equation but this time the coefficients represent moles not molecules.

Step 1: So 2 NH3 moles are formed when 3 H2 moles react.Conversion factors

2 moles NH3 or 3 moles H23 moles H2 2 moles NH3

Step 2: Create your calculation

18 moles H2 x 2 moles NH3 = 12 moles NH3 are reacted3 moles H2

**notice molecules of H2 cancel out**

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BALANCED

EQUATION

Stoichiometry:  

a quantitative study of chemical changes.

The most common type of stoichiometry calculation is a mass-mass problem. The question looks like this: "given this amount of reactant, how much product will form?" However stoichiometry calculations can involve moles, mass, gas volume and molecules.

Steps in solving a mass-mass problem:1. Write a balanced equation for the reaction. 2. Convert mass of reactant to moles of reactant. (this is what we did before)3. Convert moles of reactant to moles of product. (using balanced equation)4. Convert moles of product to grams of product. (again like before)5. Pick up the calculator and do the math.

Remember our mole conversion flow chart? Well with the balanced equation in between we now have to content with two of them. One for Products in an equation and one for Reactants. The ratios in the balanced equation act as the bridge. This will become your best friend.

x # atomsatoms molecule x N x mol mass

mol molar mass x molecule Molecules x molar mass

#atoms moles mol density x mol x mol (g/L) N 22.4L

N/mol or mol/N x 22.4L volume

mol

x # atomsatoms molecule x N x mol mass

mol molar mass x molecule Molecules x molar mass

#atoms moles mol density x mol x mol (g/L) N 22.4L

N/mol or mol/N x 22.4L volume

mol

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Let’s try one together:

If iron pyrite, FeS2, is not removed from coal, oxygen from the air will combine with both the iron and the sulfur as coal burns. If a furnace burns an amount of coal containing 100 g of FeS2, how much SO2 (an air pollutant) is produced?

4FeS2 + 11O2 2 Fe2O3 + 8 SO2

Step 1:determine starting point100g FeS2Step 2:convert to moles (using periodic table)100g FeS2 x 1 mol FeS2

120.0g FeS2Step 3:change from moles of FeS2 to moles of SO2 using ratios in the balanced equation100g FeS2 x 1 mol FeS2 x 2 mol SO2

120.0g FeS2 1 mol FeS2Step 4:convert from moles to mass (using periodic table)100g FeS2 x 1 mol FeS2 x 2 mol SO 2 x 64.0g SO2 =

120.0g FeS2 1 mol FeS2 mol SO2Step5: Double check all units cancel out except SO2 and do the calculation with your calculator.100g FeS2 x 1 mol FeS2 x 2 mol SO 2 x 64.0g SO2 = 107.0g SO2

120.0g FeS2 1 mol FeS2 mol SO2

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Stoichiometry of Reactions in Solution Stoichiometry + Molarity Up until this point we have used stoichiometry with respect to moles and mass but

I could hide the number of moles reacting in an equation by giving you the molarity. Remember concentration:

units M = mol/L ***always Liters***

So far we have tackled questions like this.A) For the given equation, how many moles of HCl are required to react with 0.011mol of NaOH?    HCl(aq)  +  NaOH(aq) ----->  NaCl(aq)  +   H2O

0.011mol NaOH x 1 mol HCl = 0.011 mol HCl1 mol NaOH

We can add another dimension to the calculations by replacing moles with molarity (M)B)What volume of 0.556 M HCl has enough hydrochloric acid to combine exactly with 24.5 mL of aqueous sodium hydroxide with a concentration of 0.458 M?  The equation for the reaction is:                      HCl(aq)  +  NaOH(aq)   NaCl(aq)  +   H2O

This is involves mL NaOH mol NaOH mol HCl mL HCl conversions.

Step 1: Start with what you know.   HCl(aq)  +  NaOH(aq)   NaCl(aq)  +   H2O0.556M 0.0245L? L 0.458M

Step 2: Find the number of moles of the known compound (NaCl)(0.0245L x 0.458M) = 0.0112 mol NaOH

Step 3: Now it looks like A) above, so we convert to HCl using the Balanced Equation0.0112 mol NaOH x 1 mol HCl = 0.0112 mol HCl

1 mol NaOH

Step 4: Double check we have answered the question. Question doesn’t ask for moles it asks for volume and has provided us with a HCl concentration so we can complete the conversion

0.0112 mol HCl x 1 L = 0.02018 L HCl0.0556mol

= 20.2 mL HCl***Get sig figs from original

numbers in questionStep 5: Make a final statementIf  we measure out 20.2 mL of the HCl solution it will exactly neutralize the NaOH. If we were to complete this titration, NaOH would be the titrant in the burette and HCl would be the solution in the Erlenmeyer.

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This process of finding out exactly how much HCl is needed to neutralize the NaOH is done experimentally by TITRATION. Equivalence point: the point at which the ratio of the moles of each species involved exactly equals the ratio of the coefficients of the species in the balanced reaction equation.

Example Problem #2   

How many millilitres of 0.114 M H2SO4 solution provide the sulphuric acid required to react with the sodium hydroxide in 32.2 mL of 0.122 M NaOH according to the following equation?                        H2SO4(aq)  +  2 NaOH(aq) ------>  Na2SO4(aq)  + 2 H2O

Again we need the mL --> mol --> mol --> mL solution. Step 1: Start with what you know.    (and double check the equation is balanced correctly!)

H2SO4(aq)  +  2 NaOH(aq) ------>  Na2SO4(aq)  + 2 H2O0.114M 0.122M? mL 0.0332L

Step 2: Find the number of moles of the known compound (NaOH)Step 3: Then used the balanced equation to convert(0.0322L x 0.122M)mole NaOH x 1 mol H2SO4 = 0.0019642 mol H2SO4

2 mol NaOH

Step 4 : Answer the actual question Again hasn’t asked for mols asked for mL

V = n =0.0019642 mol H2SO4 = 0.0172L = 17.2 mL H2SO4

M 0.114M H2SO4

Step 5: Make a final statementTherefore we need 17.2 mL of 0.114 M H2SO4 to exactly neutralize 32.3 mL of a 0.122 M solution of NaOH.

*** as you get good at this you can start to show the work in one calculation***(0.0322L x 0.122M)mol NaOH x 1 mol H2SO4 x1 L = 17.2 mL H2SO4

2 mol NaOH 0.114 mol H2SO4

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What is the Limiting Reagent?

It is simply the substance in a chemical reaction that runs out first. Back to our hamburger analogy, even if you have 10 hamburger patties but only 5 buns, you can only make 5 hamburgers. The hamburger buns are the limiting factor. It seems simple enough, but it does cause people problems. Let's try a simple example. Reactant A is a test tube. I have 20 of them.Reactant B is a stopper. I have 30 of them. Product C is a stoppered test tube.

The reaction is: A + B ---> Cor: test tube plus stopper gives stoppered test tube.

So now we let them "react." The first stopper goes in, the second goes in and so on. Step by step we use up stoppers and test tubes (the amounts go down) and make stoppered test tubes (the amount goes up). Suddenly, we run out of one of the "reactants." Which one? That's right. We run out of test tubes first. Seems obvious, doesn't it? We had 20 test tubes, but we had 30 stoppers. So when the test tubes are used up, we have 10 stoppers sitting there unused. And we also have 20 test tubes with stoppers firmly inserted.

So, which "reactant" is limiting and which is in excess? We had more than enough rubber stoppers and we ended up with 10 left-over so rubber stoppers are the excess.The 20 test tubes we started with is what determined how many stoppered test tubes we ended up this so the test tubes are the limiting reactant.

Determining Limiting Reagent/Reactant (Limiting Factor)

1. Check to be sure you have a balanced equation 2. Convert the amount of reagent one that was given into the amount of a product that you could form if that reagent was completely consumed. 3. Convert the amount of reagent two that was given into the amount of the same product that you could form if that reagent was completely consumed. 4. Determine which of the answers in step 2 and 3 produced the LEAST amount of product and that will be your limiting reagent. (Also the LEAST amount of product will be your theoretical yield)5. The other reagent is call the Excess Reagent6. To find out how much of it is in excess, calculate how much excess reagent is actually used by starting with the amount of your now determined limiting reagent.

Eg. Let's say that I was given 6.0 grams of H2 and 160 grams of O2. What is the limiting reagent? How much is in excess?

H2 + O2 -----> H2O1. Check for balance

2H2 + O2 ----> 2H2O2. Convert grams of given H2 into grams of the product H2O 6.0g H2 x 1 mol H2 x 2 mol H2O = 3.0 mol H2O

2 g H2 2 mol H23. Determine the moles of H2O formed if all 160g of O2 was consumed

160.0g O2 x 1 mol O2 x 2 mole H2O = 10.0 mol H2O32.0g O2 1 mol O2

4. Compare the answers in step 2 and 3 and the one that produced the LEAST amount of water is the limiting reagent:

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H2 produces 3.0 mol and O2 produces 10.0 mol, therefore H2 determines how much can be made,

H2 is he limiting factor

5. Now find out how much is in excessTo do this determine how much O2 is used when all 6.0g of H2 is used

up.6.0g H2 x 1 mol H 2 x 1 mol O2 x 32.0g O2 = 48 g O2 is used

2.0 g H2 2 mol H2 1 mol O2

Finally subtract what is used from what is supplied160.0g – 48 g = 112g O2 is in excess

Eg 2. If you start with 56.8g of FeCl2, 14.0g of KNO3 and 40.0g of HCl are mixed and allowed to react according to the equation 3FeCl2 + KNO3 + 4HCl 3FeCl3 + NO + 2H2O + KCl what are the limiting reagent and excess reagents? How much in excess is the excess reagent?

3FeCl2 + KNO3 + 4HCl 3FeCl3 + NO + 2H2O + KCl56.8g 14.0g 40.0g

Arbitrarily I’m choosing to convert everything to NO56.8g FeCl2 x 1 mol FeCl2 x 1 mol NO x 30.0g NO =

4.49g NO126.8g FeCl2 3 mol FeCl2 mol

14.0g KNO3 x 1 mol KNO3 x 1 mol NO x 30.0g NO =4.15g NO

101.1g 1 mol KNO3 mol

40.0g HCl x 1 mol HCl x 1 mol NO x 30.0g NO =8.22 g NO

36.5 g 4 mol HCl mol

14.0g KNO3 produces the least amount of NO therefore KNO3 is the limiting reagent and FeCl2 and HCl are in excess.

14.0 g KNO3 x 1 mol KNO3 x 3 mol FeCl2 x 126.8g FeCl2 =52.68g FeCl2

101.1g 1 mol KNO3 molTherefore 56.8g FeCl2 – 52.68g FeCl2 = 4.1g FeCl2 is excess

14.0g KNO3 x 1 mol KNO3 x 4 mol HCl x 36.5 g HCl =20.2 g HCl

101.1g 1 mol KNO3 molTherefore 40.0g HCl – 20.2 g HCl = 19.8g HCl in excess

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First the hamburger analogy

My recipe for a bacon double cheeseburger is: 1 hamburger bun 2 hamburger patties 2 slices of cheese 4 strips of bacon

Based on this recipe:1. If I have five bacon double cheeseburgers:

a. How many hamburger buns do I have? b. How many hamburger patties do I have? c. How many slices of cheese do I have? d. How many strips of bacon do I have?

2. How many bacon double cheeseburgers can you make if you start with: a. 1 buns, 2 patties, 2 slices of cheese, 4 strips of bacon b. 2 buns, 4 patties, 4 slices of cheese, 8 strips of bacon c. 1 dozen bun, 2 dozen patties, 2 dozen slices of cheese, 4 dozen strips

of bacon d. 1 mole bun, 2 mole patties, 2 mole slices of cheese, 4 mole strips of

bacon e. 10 buns, 20 patties, 20 slices of cheese, 40 strips of baconf. 2 buns, 2 patties, 4 slices of cheese, 8 strips of bacon? Which item

prevented me from making 2 hamburgers?3. If you had fixings for 100 bacon double cheeseburgers, but when you were

cooking you ruined 10 of them. What percentage of the bacon double cheeseburgers do you actually make?

Now, the chemistry problem.

NOTE: The math and the concepts are identical to the above example. The only difference is the recipe.

Here are two examples of chemical recipes: Na + Cl NaCl 1 mole of H2SO4 + 2 mole NaOH produce 1 mole Na2SO4 + 2 mole H2O

Based on the recipes above:1. If I have 1 mole of NaCl

a. How many moles of sodium do I have? b. How many moles of Chloride do I have?

2. If I want to make 5 moles of Na2SO4: a. How many moles of H2SO4 do I need? b. How many moles of NaOH do I need?

3. How much Na2SO4 can I make if I have: a. 1 mole of H2SO4 and 2 mole of NaOH b. 10 mole of H2SO4 and 20 mole of NaOH c. 0.1 mole of H2SO4 and 0.2 mole of NaOH d. 1 mole of H2SO4 and 20 mole of NaOH e. 0.42 mole of H2SO4 and 0.65 mole of NaOH f. 5 grams of H2SO4 and 5 grams of NaOHg. 2 mol H2SO4 and 2 mol NaOH? Which reactant limited my product?

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