Excess and Limiting Reactants

Balanced equations tell us the ratios of the reactants used. However, in a real life situation, some reactions require more or less reactants...
reactants that are not completely used up in a reaction are excess reagents
''                     '' are completley used up ''                       '' limiting

Example

Finding Excess/Limiting Reagents

Find the limiting reagent and the reactant in excess when 0.5 moles of Zn react completely with 0.4 moles of HCl

 Zn + 2HCl -----> ZnCl₂ + H₂

 _{Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio}

_{Zn->HCl           0.5 x 2= 1.0 mole HCl}

    There are only 0.4 moles of HCl available which is less than the required 1.0 moles.

HCl->Zn           0.4 x  1  = 0.2 mols Zn
                              2
    There are 0.5 moles of Zn available which is more than the required 0.2 moles.


    The limiting reagent is HCl, Excess reagent= Zn
    all of the 0.4 moles of HCl will be used up

    when the reaction has gone to completion there will be
    0.5 - 0.2 = 0.3 moles of Zn left over.

Example 2


Find the amount of H20 produced when 1.5g of CaCO₃ react completely with 0.73g of HCl.

CaCO₃ + 2HCl -----> CaCl₂ + CO₂ + H₂O


1.5g CaCO3 x 1  x 18gH20 = 0.27g H20
                 100g

0.73gHCl x 1  x   1molH20 x 18gH20 = 0.18gH20
              36.5g  2molHCl        


Since HCl is the limiting reactant, you can only produce 0.18g of H20.

Stoichiometry with Molarity and STP

Here’s a review of the Molarity formula if you don’t remember:

Molarity=   Moles  .

                  Volume

which can also be written as:

Volume=  Moles  .         OR          Moles= Molarity X Volume

                Molarity

Volume must be measured in Litres.

Let’s try an example:

Consider this chemical equation: NiCl3(aq) + Ca(s) → CaCl2(aq) + Ni(s)

How many grams of nickel will be formed when 110mL of 0.270M of nickel (III) chloride is reacted with sufficient calcium?

STEP 1: balance the equation

2 NiCl3(aq) + 3 Ca(s) à 3 CaCl2(aq) + 2 Ni(s)

STEP 2: make a mole map (for visual purposes)

110mL of 0.270M NiCl3 -------------------------------------> ? grams Ni

                                 -----> moles NiCl3 -----> moles Ni ----->

STEP 3: calculate

110mL = 0.110L        ßconvert mL to L

mol = M x L

0.110L NiCl3 X 0.270M NiCl3  = 0.0297mol NiCl3    ßfind mole

0,0297mol NiCl3 X  2mol Ni   = 0.0297mol Ni   ßconvert using mole ratio

                               2mol NiCl3

0.0297mol Ni X 58.7g Ni  = 1.7g Ni   ßconvert to grams

                            1mol Ni

1.7g of nickel will be formed.

Here’s another example:

How many mL of 0.200M NiCl3 are required to react with 2.10 grams of calcium?

STEP 1: chemical equation is already balanced in previous example

STEP 2:  

2.10g Ca -----------------------------------> ? mL NiCl3

        ----> moles Ca ----> moles NiCl3 ----->

STEP 3:

2.10g Ca X 1mol Ca    X  2mol NiCl3  = 0.0349mol NiCl3

                    40.1g Ca          3mol Ca

V=.   mol   .        0.0349mol NiCl3  =  0.175L NiCl3

      Molarity          0.200M NiCl3

0.175L = 175mL

175mL of NiCl3 are required.

STP: Standard Temperature and Pressure

                         22.4L

                         1mol

            ------------------------->

MOLES                                STP

           <--------------------------

                         1mol

                         22.4L

Example:

Sodium carbonate decomposes into carbon dioxide and sodium oxide. How many grams of sodium carbonate are needed to produce 11.3L of carbon dioxide measured at STP?

STEP 1: find out and balance the equation

1 Na2CO3 à 1 CO2 + 1 Na2O

STEP 2: calculate

Convert the volume of CO2 into moles using STP formula

11.3L CO2 X 1mol CO2  =  0.504mol CO2

                      22.4L CO2

Use mole ratio

0.504mol CO2 X 1mol Na2CO3  =  0.504mol Na2CO3

                                1mol CO2

Convert moles to grams

0.504mol Na2CO3 X 106g Na2CO3  =  53.5g Na2CO3

                                  1mol Na2CO3

53.5 grams of Na2CO3 are needed.

Here are a few worksheets on stoichiometry and molarity:

http://misterguch.brinkster.net/PRA048.pdf

Stoichiometry Calculations

Solving Stoichiometry calculations:

1. Balance the equation (If not already)
2. Convert units of a given substance to moles.
3. Using the mole ratio, calculate the moles of substance yielded by the reaction.
4. Convert moles of wanted substance to desired units.

Mole Map for reference:

Mole Map

Example:  Zn + 2HCl --> ZnCl₂ + H₂

• How many grams of ZnCl₂ will be produced in the reaction of HCl with 0.350 moles of Zn?

0.350 mol Zn x 1 ZnCl₂/1 Zn = 0.350 mol ZnCl₂

0.350 mol ZnCl₂ x 135.4g/1 mol = 47.39 g ZnCl₂

                                         Sig Fig = 47.4 g

Example: Ca(OH)₂ + H₂SO₃ --> CaSO₃ + 2H₂O

• How many particles of water can be produced from 47.6 g of Sulphurous acid (H₂SO₃) when sufficient Calcium hydroxide (Ca(OH)₂) is available?

47.6 g H₂SO₃ x 1 mol/82.1 g = 0.57978... mol H₂SO₃

0.57978 mol H₂SO₃ x 2 H₂O/1 H₂SO₃ = 1.15956... mol H₂O

1.15956 mol H₂O x 6.022x10²³/1 mol = 6.9828794x10²³ particles of H₂O

                                                 Sig Fig = 6.98x10²³ particles of H₂O

Example: 2Al +3Cl₂→2AlCl₃

•  80.0 g of Al is reacted with excess chlorine gas, how many formula units of AlCl₃ are produced?

80g Al x 1 mol/27g = 2.96 mol Al

2.96 mol Al x 2 AlCl₃/2 Al = 2.96 mol AlCl₃

2.96 mol AlCl₃ x 6.022x10²³/1 mol = 1.78x10²⁴ 

                                             Sig Fig = 1.8x10²⁴ formula units of AlCl₃

Example: CO +2H₂→CH₃OH

• At STP, what volume of H₂ is needed to react completely with 8.02×10²³molecules of CO?

8.02×10²³molecules CO x 1 mol/6.022x10²³ = 1.3321... mol CO

1.3321 mol CO x 2 H₂/1 CO = 2.66423... mol H₂

2.66423 mol H₂ x 22.4 L STP/1 mol = 59.6787 L H₂

                                               Sig Fig = 59.7 L H₂

Example: Cu + 2AgNO₃ --> Cu(NO₃)₂ + 2Ag

• How many g of Ag are formed when 16.2 g Cu react with sufficient AgNO₃?

16.2g Cu x 1 mol/63.5g = 0.255 mol Cu

0.255 mol Cu x 2 Ag/1 Cu = 0.510 mol Ag

0.510 mol Ag x 107.9 g/1 mol = 55.054 g 

                                     Sig Fig = 55.1 g of Ag

• How many atoms of Ag are formed when 2.5 mol Cu are reacted with sufficient AgNO₃?

2.5 mol Cu x 2 Ag/1 Cu = 5 mol Ag

5 mol Ag x 6.022x10²³/1 mol = 3.011x10²⁴

                                   Sig Fig = 3.0x10²⁴ atoms Ag

Stoichiometry

is a branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions.



Stoichiometry Calculations
step 1. Write a balanced equation
step 2. Map the solution
step 3. Calculation


Mole-Mass Calculations
A chemical equation is written in terms of moles of reactants and products. However, in the lab, substances are measured in units of mass. So you will need to translate the mole to mass or mass to mole.


Example
Zn + 2 HCl --> ZnCl2 + H2
a)How many grams of Zn will react with 1.05 moles of HCl?
step 1: Zn + 2 HCl --> ZnCl2 + H2
step 2:   1.05 moles of HCl ------------------------> grams of Zn
                                                   moles of Zn
step 3:  1.05molHCl x 1molZn/2molHCl x 65.4gZn/mole = 34.3g Zn


b) How many moles of HCl are needed to form 6.12 grams of ZnCl2?
step 1:  Zn + 2 HCl --> ZnCl2 + H2
step 2: 6.12 grams of ZnCl2---------------------->moles of HCl
                                                 moles of ZnCl2
step 3: 6.12g ZnCl2 x 1mole/136.4g ZnCl2 x 2mol HCl/1mol ZnCl2 = 0.0897 mol HCl


Mass-Mass Calculations
When you calculate the number of grams of product yielded by a certain mass of reactant.


Example video:
<iframe title="YouTube video player" width="480" height="390" src="http://www.youtube.com/embed/tr_mJnrTp2s" frameborder="0" allowfullscreen></iframe>
<iframe title="YouTube video player" width="480" height="390" src="http://www.youtube.com/embed/y3SstUNwEHk" frameborder="0" allowfullscreen></iframe>


Molarity and Stoichiometry
molarity (M) = moles of solute/liter of solution (mole/L)



Example:
How many grams of Cu will be formed when 120mL of 0.3M CuSO4 is reacted with sufficient Zn?
Step 1:  1CuSO4 + 1Zn -------> 1ZnSO4 + 1Cu
Step 2: 120mL of 0.3 M CuSO4-------------------------->grams of Cu
                                             moles CuSO4---> moles Cu
Step 3: 0.12L x 0.3 mol CuSO4/1L x 1mol Cu/1mol CuSO4 x 63.5gCu/mole = 2.29g Cu