Limiting Reagent

In every reaction, reactants are not always present in exact stoichiometric amounts.  There  is always reactant that exceeds the required amount and is called excess reagent and the reactant that limits the reaction is called limiting reagent.   

Just like when you want to bake a cake and you buy the ingredients.  Not all ingredients you bought will be used in baking because it will depend on the required amount.  Let see for example you need to use 2 cups of sugar and 4 cups of flour and the available ingredients are 4 cups both ingredients, what is the limiting reagent and the excess reagent.  If we will try to analyze,

        given:     4 cups Sugar
                       4 cups flour

     needed:     2 cups sugar
                       4 cups flour

If we will use all 4 cups of sugar, we need 8 cups of flour and we don't have enough flour to use.  But if we will follow the amount of flour, 4 cups of flour needs only  2 cups of sugar and we still have an excess of 2 cups, therefore the excess reagent is the sugar and the limiting reagent is the flour.  Flour will be the one that will limit the reaction.

Let us apply this in an example:

The reaction between aluminum and iron (III) oxide can generate temperature approaching 3,000 C  and is used in welding metals:

In one process, 120 g of Al are reacted with 600 g of Fe2O3.

a) Calculate the mass in grams of Al2O3 formed.

b) How much of the excess reagent is left at the end of the reaction?

Solution:

In order to determine the limiting reagent, we need to calculate the amount of Fe2O3 if 200 g of Al is used, and calculate the amount of Al if 600 g of Fe2O3 is used.

 From the calculation above, Al is the limiting reagent and the excess reagent is the Fe2O3.

TRY THIS:

1. Nitric oxide (NO) reacts with oxygen gas to form nitrogen dioxide (NO2) a dark brown gas:

In one experiment  0.885 mole of NO is mixed with 0.500 mole of O2.  Calculate which of the two reactants is the limiting reagent.  Calculate also the nunber of moles of NO2 produced.

2.  Urea [NH2)2CO] is prepared by reacting ammonia with carbon dioxide:

 In one process, 637 g of NH3 are treated with 1142 g CO2.

 a.  Which of the two reactants is the limiting reagent?
 b.  Calculate the mass of (NH2)2CO formed.

 c.  How much excess reagent (in grams) is left at the end of the reaction?

 

Stoichiometric Calculation of Chemical Equation

Stoichiometry is a quantitative study of reactants and products in chemical reaction.  Stoichiometry calculation of chemical reaction enables us to predict the amount of products that can be produced from the given reactants and vice versa.

In order to solve stoichiometric problems of chemical equation you need to know what does a balance chemical equation mean.  Let us have an example in the combustion reaction of  carbon monoxide.

The equation above means that 2 moles of CO reacts with 1 mole of O2 to produce 2 moles of CO2.  It can also be interpreted by using the number of molecules.  This also means that when 2 molecules of CO react with one molecule of O2, there are 2 molecules of CO2  produced.  

Based from the given balanced equation you can already predict the amount of either products or reactants depending on the given.  For example, if there are 3 moles of CO how much O2 is needed to produced CO2?  

Now to solve the problem you have to identify the relationship that you can use in solving the problem.   Since the question is asking the amount of O2 given the CO, you need to determine the ratio between O2 and CO.

To solve for the problem above we will use the relationship that 2 moles of CO needs 1 mole of O2 to produced  CO2.  There is 2:1 ratio between CO to O2.  Using the factor label method in solving the problem:

What about the amount of moles of CO2 produced?  You will now use the relationship either CO to CO2 or O2 to CO2.   The ratio of CO to CO2 in balanced equation is 2:2 therefore, 3 moles of CO will also yield 3 moles of CO2. 

The problem above is an example of mole to mole calculation.  A calculation wherein the given is number of mole and asking the number of mole also.

Another type of stoichiometric  calculation is mass to mass calculation.  In this type of stoichiometry the given is mass and what is unknown is also mass.  For example, using the same equation above, the reaction between CO to O2 to form CO2.  How much CO2 will be produced when 10.5 g of CO is added to O2 to form CO2?  So the above problem needs first the conversion of mass to no. moles and vice versa which means that the molar mass of the compound will be used.  

The above conversion step can be used to solve the problem.  Since the mass of CO is given, we need to calculate first the no. of moles of CO, and mole to mole conversion of CO to CO2 and grams or mass of CO2 can be calculated using the molar mass of the CO2.  

Next is to solve the moles of CO2

And to calculate the mass of CO2, we need to use the molar mass of CO2 which is 44 g.

So the mass of CO2 is 16.6 g that can be produced from 10.5 g of CO.

We can combine the three calculation in one step only, 

Sample Problem 1

If 200 g of methanol are used up in a combustion process , what is the mass of CO2 produced?

The balanced equation of the reaction of methanol with oxygen is 

Solution:

In solving the problem we need to follow several steps.  

Step 1.  Convert mass of methanol to no. of moles.  To convert mass to no. of  moles,  molar mass of methanol should be calculated first.  Molar mass of CH3OH = 12 g + 4(1 g) + 16 g =  32 g/mol.  (Note: In this blog the atomic mass used are always in whole number, meaning it is already rounded off).

Step 2.  Calculate no. of mole of CO2 using the calculated mole of CH3OH using the balanced equation.  From the balanced equation, in every 2 moles of CH3OH yields 2 moles of CO2.  (2:2 mole ratio)

Step 3.  Calculate the mass of CO2 using the mole of CO2 calculated above.  Molar mass of CO2 is needed.  Molar mass of CO2 = 12 + 2(16) = 44 g/mol.  

The answer is 275 g of CO2.

Now if you want to have shorter solution, we can combine all steps

TRY THIS:

1.  Limestone (CaCO3) is decomposed by heating to quicklime (CaO) and carbon dioxide.  Calculate how many grams of quicklime can be produced from 350 g of  limestone.

2. Nitrous oxide (N2O) is also called laughing gas.  It can be prepared by the thermal decomposition of ammonium nitrate (NH4NO3).  The other product is H2O.

 a)  Write a balanced equation for this reaction.

 b)  How many grams of N2O are formed if 0.46 mole of  NH4NO3 is used in the reaction?

Double Replacement Reaction

Double Replacement Reaction is a type of chemical reaction where two ionic compounds exchange anions to produce two new compounds. The general form of double replacement reaction is

The general  form above shows the exchange of ions of the two compounds AX and BZ where A paired with Z forming AZ and B paired with X forming BX..  It is understood that A is a positive ions and should be paired with Z which is negatively charged ions.  Same with the second compound formed B is positive ion and should be paired with the negative ions X.

The second equation above shows that there is an exchange of ions between compounds in aqueous solution. Aqueous solution means that the compound is dissolved in water. Reaction comes into completion only when there is a formation of precipitate or a formation of  insoluble salts as shown in the equation above.  But if both compounds are soluble in water then there is no reaction, meaning both products formed aqueous solution.

In order to predict if an ionic compound is soluble or insoluble use Solubility Rules.

Example:

The example above shows that reaction is possible due to the formation of insoluble precipitate 

Observe the next example:

The above example shows that both possible products are soluble in water, therefore there is no reaction between the two reactants.  Final answer will be: 

Neutralization Reaction is a special kind of double replacement reaction.  This reaction involves the reaction between an acid and a base forming a salt and water.

Example:  

TRY THIS:

FOR ANSWERS' KEY CLICK HERE

Single Replacement Reaction

Single replacement reaction involves the reaction either of a metal with an aqueous solution or a metal with an acid aqueous solution..  It is called replacement reaction since the more active element replaces a less active element from a solution.

1.  Reaction of Metal with an Aqueous solution

     Example:
   

   
In the above equation  Zn metal is added to aqueous silver nitrate solution.  To predict whether reaction is possible or not, activity series of metals   will be used.  Since the position of Zn is higher in the activity series of metal than Ag, it means that Zn is more active than Ag, therefore Zn replaces Ag in the equation forming aqueous zinc nitrate solution.  But if we will have the reverse reaction,

If we do the reverse reaction therefore no reaction will occur, considering that Ag is lower in position in the activity series, meaning Ag is less active than Zn and therefore Ag cannot replace the Zn.

2. Reaction of metal with Aqueous acid solution

     Example:

In the above equation Ni replaced H in the reaction.  Since the location of Ni in the activity series of metal is higher that H, therefore Ni is more active than H and so reaction occurs forming aqueous nickel (II) chloride and hydrogen gas.

Watch video on Single Displacement Activity in YOUTUBE

TRY THIS:

CLICK HERE  FOR ANSWER'S KEY

Writing and Balancing Chemical Equation

In writing chemical equation, you must have knowledge in writing chemical formulas.  Chemical formulas, symbols, arrow, + sign are part of chemical equation.  Chemical equation is a shortcut way of writing chemical reaction.  It gives you the reactants and the products.  Reactants are the substances that undergo reaction and the products are the result of the reaction.  Reactants are written in the left side of the equation and the products are written in the right side of the equation separated by an arrow.

  

Chemical equation also shows the phases of the reactants and products.

          s or arrow downward stands for solid

          l stands for liquid

          g or arrow upward stands for gas

          aq stands for salts or substance dissolved in water

But I want to emphasize how to write and balance chemical equation and so sample equations will not show the phases of the reactants and products.

HOW TO WRITE CHEMICAL EQUATION

When you write chemical equation make sure that it tells the truth, meaning correct symbols and formulas must be written.  You must be familiar with the different diatomic molecules, they are useful in writing chemical equations.

Example:  

1.  Carbon reacts with oxygen gas to form carbon dioxide.

Answer:

            

Carbon will be written in elemental form while oxygen as diatomic molecule. Carbon dioxide will have 1 atom of carbon and 2 atoms of oxygen.  It is understood that the oxidation of carbon used is 4 and oxygen is two.

2.  Sodium and oxygen react to form sodium oxide.

Answer:  

Sodium also will be in elemental form and oxygen will have a subscript of two since a diatomic molecule.  The formula of sodium oxide is, sodium having 2 subscript from the oxidation of oxygen and oxygen has subscript of 1 from the oxidation of sodium.  

As you can see the equation gives the correct formulas and symbols needed in the equation.  But if you will check the equation is not yet balanced.  Equation must be balanced to conform with the Law of Conservation of Mass.  According to the law the total mass of the reactants must be equal to the total mass of the products.  This can only be true if the equation is balanced.  

HOW TO BALANCE CHEMICAL EQUATION

In balancing chemical equation, you just need to follow several steps.  Below is the simplified way of balancing chemical equation.  

1.  First, write correctly the symbols and formulas of reactants and products.  And then count the number of atoms of the elements in the reactants side and in the products side. 

Example :  

1.  Write the balanced equation of the reaction between sodium  and oxygen  forming sodium oxide.

Reactant side                    Product side

Na  =  1                             Na  =  2

O    =  2                             O    =  1

Notice the number of atoms, sodium in the reactant side is only one but in the product side is 2 and oxygen in the reactant side is two while in the product side is only one.  This only indicates that the equation is not balanced.

2.  Balance the equation by writing coefficient written before the formula.  Coefficient must be a whole number and must be in a smallest ratio.   Subscript must not be change, if you do so, your changing the symbols and formulas of the elements and compounds in the equation.  This is a big NO NO in balancing equation.

From the above equation, we can balance it by writing coefficient 2 before sodium oxide to balance the oxygen and write 4 before the Na to balance the sodium atom.  

And to check if balance

       Na  =  4                                             Na  =   4

       O    =  2                                             O    =  2

Therefore the equation is already balanced.

Other examples:

1.  Write the balanced equation of the reaction between phosphorous and oxygen to form                           diphosphorous pentaoxide.

2.  Write the balanced equation between the reaction of Iron and chlorine to form Iron (III) chloride.

Solution:

1.  The equation looks like this:

     To balance, simply write 5 before oxygen and 2 before diphosphorous pentaoxide to balance the          oxygen, and to balance phosphorous write 4 before phosphorous.

2.  Equation in number 2 is:

     Balance by writing 3 before chlorine gas and 2 before Iron (III) chloride to balance the chlorine          and to balance Iron write 2 before Iron.

     That makes the Iron 6 and Iron 2, therefore the equation is balanced.

Solubility Rules

Solubility is a property of a substance that determines the ability of the solutes to dissolve in a given solvent.  Solubility Rules is used to predict whether a particular salt is soluble or not soluble in water.  This can be used also in predicting the possible products in double replacement reaction.

Some Common Monoatomic and Polyatomic Ions

In writing chemical formulas you need to familiarize the different monoatomic and polyatomic ions.
Below is provided for you: (Click here for pdf copy)

Activity Series of Metals including Hydrogen

Activity Series of Metals is used to predict whether a reaction is possible or not in single replacement reaction.  The metals including hydrogen are arranged from the most reactive metal down to least reactive metal.  A metal can only replace an element below it.

Li           Lithium          Most Active Metal
Rb          Rubidium        
K            Potassium
Cs          Cesium
Ba          Barium
Sr           Strontium
Ca          Calcium
Na          Sodium
Mg         Magnesium
Al           Aluminum
Mn         Manganese
Zn          Zinc
Cr           Chromium
Fe           Iron
Ni          Nickel
Sn          Tin
Pb          Lead
H  -        Hydrogen
Cu          Copper
Hg          Mecury
Ag          Silver
Pt            Platinum
Au          Gold          Least Active Metal


Based from the Activity Series of metals Lithium is the most active metal while Gold is the least active metal.  Metal can only displace or replace metal below it.  Example, metals above hydrogen are the metals that can displace hydrogen and the metals below hydrogen.  While metals below hydrogen are the metals that hydrogen can displace.

ACTIVITY SERIES OF NONMETALS

F

Cl

Br

O

I

S

The activity series of nonmetals is arranged from the most active to the least active nonmetal.  Flourine is the most active nonmetal and sulfur is the least active nonmetal.

Simple Rules in Naming Chemical Compounds

A. Binary Compounds of Metal and Nonmetal
     This compound is composed of metal and nonmetal. If you want to learn how to name this compound just follow the simple rule below:

     Rule:  Name of metal  + Stem name of nonmetal + -ide ending
   
     Example:
          NaCl    =          Na           +        Cl
                       =      Sodium       +       Chlor (stem name of chlorine) +  ide
                       =      Sodium Chloride

     Some more examples:
         KCl       =   Potassium chloride
         MgO     =   Magnesium oxide
         Li2S      =   Lithium sulfide
         MgCl2  =   Magnesium chloride
         BeO     =   Berrylium oxide

B.  Binary Compounds of Nonmetal and Nonmetal
      This compound is composed of two nonmetals.  Now in naming this compounds prefixes are used.

 Prefixes

     1     -   mono
     2     -   di
     3     -   tri
     4     -   tetra
     5     -   pent
     6     -   hex
     7     -   hept
     8     -   oct
     9     -   nona
    10    -   dec

     Rule:  Prefix of first element except mono +  Name of first nonmetal + Prefix of second nonmetal + stem name of second nonmetal + ide

     Example: 

     CO =      C         +          O
           =    Carbon (1 carbon no prefix since mono)  +  Mono (1 oxygen ) + ox (stem name of oxygen) + ide 
           =    Carbon monoxide

     Other examples:
      
      CO2    -   Carbon dioxide
      NO3    -   Nitrogen trioxide
      N2O4  -   Dinitrogen tetraoxide
      CCl4   -   Carbon tetrachloride
      P2O4   -   Diphosphorus tetraoxide

C.  Binary Acids 
     This compound is composed of hydrogen bonded to a nonmetal

     Rule :  Hydro (constant prefix) + stem name of nonmetal + -ic acid

     Example :
 
    HCl   =          H          +         Cl
              =     Hydro   +   chlor  (stem name of chlorine) +  ic acid
              =    Hydrochloric acid

     Other examples;
     H2S    -   Hydrosulfuric acid  ( sulfur is exempted to the rule, full name is used)
     HF      -   Hydroflouric acid
     HBr    -   Hydrobromic acid


D.  Binary compounds of metal with variable oxidation number and nonmetal
      This compound is composed of metal having two or more oxidation number paired with a nonmetal.

     There are two rules to follow, one is by the use of IUPAC system of naming and the latin system of naming.

IUPAC System:

     Rule :  Name of metal + Roman Numeral showing the oxidation used in metal element +  Stem name of nonmetal  +  ide

     Example :

     FeCl2   =     Fe (+2 oxidation used)      +   Cl  (-1 oxidation used)
                 =     Iron (II) (II - indicates the oxidation number used of Fe)  +  Chlor (Stem name of chlorine + ide
                 =     Iron (II) chloride

     Other examples:

     FeCl3    -  Iron (III) chloride
     Fe2O3    -  Iron (III) oxide
     FeO      -  Iron (II) oxide

  Latin System:

     Rule :  Stem name of the latin name of metal ( but there are some exemption like Hg)  + ic ( is attached when the oxidation number used of the metal is higher) or  -ous (is attached when the oxidation number used of the metal is lower) + stem name of nonmetal + ide

     Example:

     FeCl2   =     Fe (+2 oxidation used)    +      Cl   (-1  oxidation used)
                  =    Ferr  ( Stem name of the latin name of Fe which is Ferrum)  +  ous ( ous is the suffix used for lower oxidation number  +  chlor (stem name of chlorine)  +  ide
                  =    Ferrous chloride
 
     Other examples:
   
     FeCl3     -   Ferric chloride
     Fe2O3     -   Ferric oxide
     FeO        -  Ferrous oxide

E.  Ternary Ionic Compounds
      This compound is composed of metal attached to a  polyatomic anion.

     Rule:   Name of metal ( use roman numeral of the oxidation number when metal has variable oxidation number +  name of the polyatomic anion.

     Example :

     NaOH   =     Na        +          OH
                  =    Sodium (the name of the metal)  + hydroxide ( name of the polyatomic anion)
                  =    Sodium hydroxide

     Other Examples:

     CaSO4        -  Calcium Sulfate
     Cu3PO4      -  Copper (I) phosphate
     Cu3(PO3)2 -  Copper (II)  phosphite


F. Ternary Acids (Oxyacids)
     Ternary acids are acids composed of hydrogen element paired to polyatomic anion.  This is also called oxyacids, acids containing oxygen.

     Rule :  Stem name of the element which is not hydrogen and oxygen (but there are some exemption, other element use full name like sulfur, carbon) + ous acid ( this is used when the ending of the polyatomic anion is ite) or ic acid (when the ending name of the polyatomic anion is ate).

     Example

     H2SO4   =       H      +    SO4  (sulfate, name of polyatomic anion)
                  =     Sulfur    +   ic acid ( ic acid is used because sulfate ends in ate)

     Other examples:

     H3PO3    -   Phosphorous acid
     H3PO4    -   Phosphoric acid
     H2SO3    -   Sulfurous acid


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Concentration of Solution Sample Problems

A. Percent by Mass Sample Problems

     1.  What is percent by mass concentration of 12.5 g of NaCl dissolved in 200 g of water?

          Identify the given:   mass solute = 12.5 g NaCl
                                       mass solvent = 200 g water

          The problem is asking the percentage by mass of NaCl given the mass of solute and the mass of solution.  You have to take note that the mass of the solution is equal to the mass of solute + the mass of solvent.

          Formula:   

 

          Solution:

 

 

     2.   Calculate the percentage by mass of 30 g of KCl in 250 g of solution.

          Find the given:    mass of solute = 30 g KCl

                                    mass of solution = 250 g of solution

         Here the problem is asking to calculate the percentage by mass given the mass of the solute and the mass of the solution.

          Formula:

          Solution:

 

     3.  What is the mass of the CuSO4 present in 15% by mass 316 g of CuSO4 solution?

          

          Find the given:   Percent by mass = 15 %

                                   mass of solution =  316 g of solution

         The problem asked to calculate the mass of CuSO4, which is the solute of the solution.

          Formula : 

          Solution:
 

B. Percent by Volume Sample Problems

1.  A solution of hydrochloric acid is produced by dissolving 15 mL of HCl in enough water to make 240 mL of solution. What is the percentage composition of the solution?

     Find the given :    volume of solute = 15 mL

                                volume of solution = 240 mL

     This problem is asking the percentage  by volume of hydrochloric acid given the volume of solute and the volume of solution.

     Formula :

 

     Solution:

 

2.  Calculate the volume of alcohol present in 250 mL 70% alcohol solution and  40 % alcohol solution.

        Find the given:   volume of alcohol solution = 250 mL

                                 Percentage by Volume = 70 % and 40 %

        The problem is asking the volume of alcohol present in 70 % and 40% alcohol solution having 250 total volume of solution.  Two solutions are needed in this problem.

         Formula: 

          Solution:

     

          For 70% alcohol solution

 

          For 40% alcohol solution

 

 

     3. The average adult human body contains about 5 L of blood.  Only 0.72% consists of leukocytes (white blood cells).  What volume of leukocyte cells is present in the body of a small child, with only 2.5 L of blood?

         The given in this problem is the percent of white blood cells = 0.72% and the total volume of blood of a child blood = 2.5 L.

          The problem is asking to calculate the volume of leukocytes or the white blood cells.

         

         Formula:  

          Solution: 

C. Molarity Problems

1.  What is the molar concentration of the following solutions:

     a. 0.20 mol of NaCl in 0.25 L of solution.

     b.  5 g of sugar, C12H22O11 in 20 mL of solution.

     c.  3 g of KCl in 80 mL solution.

Formulas needed:

 

 Solutions:

     a.  Given :  mole solute = 0.20 mol

                      volume of solution = 0.25 L of solution

 

          Unit for Molarity can be mol/L or simply M.

     b.  Given:  mass of the substance = 5 g

                     volume of solution = 20 mL or 0.02 L

                     

           In this problem moles of solute should be calculated first, and needs the molar mass of sugar

    

           Molar mass of sugar = 12 (Atomic Mass of Carbon) + 22 (Atomic Mass of Hydrogen) + 11(Atomic Mass of Oxygen)

           Molar Mass of Sugar = 12(12) +22(1) + 11(16)

                                            =  144 + 22 + 176

                                            =  342 g/mol

           How to calculate mole of solute?

          Calculation of Molarity:

        

 

      c.  Given:  mass of the substance = 3 g of KCl 

                      volume of solution = 80 mL solution or 0.08 L 

          Molar mass of KCl = 1(atomic mass of K) + 1(Atomic Mass)

                                        =  1(39) + 1(35)

                                        =  39 + 35

                                        =  74 g/mol

         Calculation for mole of solute

 

         Calculation for molarity:

2.  What mass of solute is present in each of the following solutions?

      a.  400 mL of 0.05 M of  Calcium hydroxide, (Ca(OH)2) solution.

      b.  2.5 L of 1.00 M of sodium chloride, (NaCl) solution

       Solution:

      a.  Given :  Volume of solution = 400 mL or 0.4 L

                       molarity = 0.05 M or mol/L

          In this problem we need to calculate first the mole of the solute before we can calculate the mass.

        Calculating the moles of solute, formula is just rearrange:

   
         Calculating for the mass of solute, needs the molar mass of Ca(OH)2

          Molar Mass of Ca(OH)2 = 1(Atomic Mass Ca) + 2 (Atomic Mass O) + 2(atomic Mass H)

                                                 = 1(40g)  +  2(16g) + 2(1g)

                                                 =  40g + 32g + 2g

                                                  =  74 g/mol

           Calculating mass of the substance:

          

     b.  Given : volume of solution = 2.5 L
                      Molarity =  1 M of NaCl solution


         
            Before the mass can be calculated the mole of solute should be calculated first and the molar mass of the substance

            Molar Mass NaCl = 1 (23) + 1 (35)
                                         =  23 + 35
                                          =  58 g/mol

             Calculating the mole of solute:

                           

             Calculating mass of substance: