March 12th- Excess & Limiting Reagents

Today we learned about  EXCESS & LIMITING reactants.

Chemical reaction equations give the ideal stoichiometric relationship among reactants and products.
However, the reactants for a reaction in an experiment are not necessarily a stoichiometric mixture. In a chemical reaction, reactants that are not use up when the reaction is finished are called excess reagents. The reagent that is completely used up or reacted is called the limiting reagent, because its quantity limit the amount of products formed.

Example 1 

using the following equation, find the excess reagent when 10.0 g manganese metal is allowed to react with 10.0g nitric acid

first make the formula

3Mn + 8HNO3----- 3Mn(NO3)2 + 2NO + 3H2O

then you find the mass of HNO3 using 10g of manganese(order doesnt matter)

10g x 1moleMn/54.9 x 8moleNO3/3moleMn x 63gHNO3/mol HNO3
=30.6g of HNO3 .......since you only need 10g of HNO3, but you have 30.6g of HNO3 it is excess
it also means  Mn is limiting

Example 2 

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

1. Write the balanced chemical equation for the chemical reaction     Zn + 2HCl -----> ZnCl₂ + H₂
    
2. Calculate the available moles of each reactant in the chemical reaction

   moles of Zn = 0.5

   moles of HCl = 0.4

3. Use the balanced chemical equation to determine the mole ratio of the reactants in the chemical reaction

   Zn : HCl	Or	HCl : Zn
   1 : 2		1 : ½

4.  Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio     If all of the 0.5 moles of Zn were to be used in the reaction it would require
       2 x 0.5 = 1.0 moles of HCl for the reaction to go to completion.
       There are only 0.4 moles of HCl available which is less than the required 1.0 moles.
       If all of the 0.4 moles of HCl were to be used in the reaction it would require
       ½ x 0.4 = 0.2 moles Zn.
       There are 0.5 moles of Zn available which is more than the required 0.2 moles
    .
5. The limiting reagent is the reactant that will be completely used up during the chemical reaction.
       There will be some moles of the reactant in excess left over after the reaction has gone to completion.     The limiting reagent is HCl,
       all of the 0.4 moles of HCl will be used up when this reaction goes to completion.
       The reactant in excess is Zn,
       when the reaction has gone to completion there will be
       0.5 - 0.2 = 0.3 moles of Zn left over.
   
   

Here are more practice questions!

http://www.ausetute.com.au/exceslim.html

here is the website that will help you
http://www.youtube.com/watch?v=Vaiz0zLesHk&feature=related

March 7- Stoichiometry Calculation

Today we did stoichiometry calculations involving Molarity and STP.

Molarity is easily defined by this one equation. Molarity=  Moles
                                                                                   Litres

STP is 22.4 litres/mole

Example

1. How many moles of nitrogen gas is needed to react with 44.8 liters of hydrogen gas to produce ammonia gas?

3H₂   +   N₂    2NH₃
GIVEN: 44.8 L of H₂ at STP.
FIND: mols of N₂.
Here the sequence is: GIVEN liters of H₂ at STP, CHANGE liters of H₂ at STP to mols of H₂, MOL RATIO to change from H₂ to N₂. There is no need to go any further to change the N₂ into mols, because the mol ratio leaves the material in that unit anyway.






How many liters of ammonia are produced when 89.6 liters of hydrogen are used in the above reaction?
The "above reaction" from problem #1 is: N₂   +   3H₂    2NH₃

GIVEN: 89.6 L of H₂ at STP.
FIND: Volume of ammonia (in liters at STP)
Take the GIVEN quantity, use the Molar Volume of Gas at STP (MVG) to change it to mols, change the material with the mol ratio (MR), and change the mols of new material to the requested liters at STP using the MVG again.

March 5th - Nick kim

 Today we have learned how to do Stoichiometry calculation involving particles-moles-mass
we have acutually covered how to do conversion between particles-moles-mass, thus we just need to know how are we gonna use stoichiometry to find answers.
 [Ex] Cu+2AgNO3----Cu(NO3)2+2Ag

How many grams of AgNO3 are required to completely react with 2.1 moles of Cu

First we should find moles of AgNO3 in order to find its weight. Thus, we use what we learned last class Stoichiometry( using ratio to find the number of moles)

2.1moles of Cu x 2moles of AgNO3/1mole of Cu-------- If you look above then you see ratio between Cu and AgNO3 is 1:2 right so you put that into fraction and make unwanted element cancel each other.

Therefore we end of having 4.2 moles of AgNO3-------- and now we need to convert this into grams, u find the mass of AgNO3 from periodic table which is 107.9g+14g+16(3)g=169.9g
Then we do mole conversion.

4.2 moles of AgNO3 x 169.9g of AgNO3/ 1 mole of AgNO3------ moles cancel each other and left with multipication.

=713.58(careful about the sigfig)

Here is the diagram that will help.

More practices

http://www.standnes.no/chemix/examples/stoichiometry-problems-chemistry.htm

video that will help us to understand more

http://www.youtube.com/watch?v=49k88Alfh88

march 1 marcus lu

stoichiometry!

Stochio is Greek for element, and metry means measurement.  Therefore, stoichiometry is the branch of chemistry that deals with quantities of elements and compounds involved in a reaction.


Using the example 4NH3 + 5O2 --> 6H20 + 4NO, we can conclude that the mole ratio is 4:5:6:4.

4 moles of NH3 react with 5 moles of O2 to produce 6 moles of H20 and 4 moles of NO.

AN example question is : How many moles of O2 will be formed when 9000 moles of H20 is decomposed?

9000mole H20 X 1mole O2 / 2mole H20 = 4500 mol O2

**** we divide the 9000moles of H20 by 1/2 because in H20, there is 2 moles of H and 1 mole of O!