Title: Limiting
1Limiting Excess Reactants
- How do you know which one is which?
2What does limiting excess mean?
- Limiting Reactant - the reactant that runs out
first in a chemical reaction, thus determining
the amount of product produced - Excess Reactant - the reactant that there is a
quantity of left over after a chemical reaction - The excess reactant should be the cheaper
reactant since we do not like to waste unused
chemical - The excess reactant should be easy to isolate
from the product(s) once the reaction is complete
since we do not want contamination
3Vinegar (5 by mass acetic acid) and Baking Soda
Demo
- NaHCO3 (s) HC2H3O2 (aq) ? NaC2H3O2 (aq) H2O
(l) CO2 (g)
Flask 1 Flask 2 Flask 3
Mass of Baking Soda 1.50 g 3.50 g 5.50 g
Mass of Vinegar 50.00 g (2.5 g HC2H3O2) 50.00 g (2.5 g HC2H3O2) 50.00 g (2.5 g HC2H3O2)
4Outcome from Demo
- Flask 1 showed the smallest amount of CO2
produced in the balloon with a clear solution in
the flask. - Flask 2 showed the same amount of CO2 produced
in the balloon as in flask 3, which was a larger
amount compared to flask 1. The solution in the
flask was still clear like in flask 1. - Flask 3 showed the same amount of CO2 produced
in the balloon as in flask 2. However, the
solution in the flask was cloudy.
5Which reactant was limiting and which reactant
was excess in each flask based off observations?
Flask 1 Flask 2 Flask 3
Baking Soda limiting theoretical amount excess
Vinegar excess theoretical amount limiting
6The proof is in the stoichiometry
50.0 g vinegar 5 g HC2H3O2 1 mol HC2H3O2 1 mol NaHCO3 84.01 g NaHCO3 3.50 g NaHCO3
100 g vinegar 60.06 g HC2H3O2 1 mol HC2H3O2 1 mol NaHCO3
The theoretical amount of baking soda needed to
react completely with the 50.0 grams of vinegar
is 3.50 g.
7Practice Problem
- 2NaI Cl2 ? 2NaCl I2
- 1. You are given 22.1 g of NaI and 4.13 g of
Cl2. What is the limiting reactant? - Pick one of the values given for your reactants
and through stoichiometry find out how much you
need of the other reactant.
822.1 g NaI 1 mol NaI 1 mol Cl2 70.90 g Cl2 5.23 g Cl2
149.89 g NaI 2 mol NaI 1 mol Cl2
You have 4.13 g Cl2 available You need 5.23 g to
react completely with 22.1 g NaI Therefore, Cl2
is the limiting reactant which will run out first
in the reaction and determine the amount of
product produced.
9- How much NaCl is produced?
- Remember that you must always start with the
limiting reactant quantity that you have because
this is what determines the amount of product
produced.
4.13 g Cl2 1 mol Cl2 2 mol NaCl 58.44 g NaCl 6.81 g NaCl
70.90 g Cl2 1 mol Cl2 1 mol NaCl
10- How much excess reactant will be left over?
- Remember that you must always start with the
limiting reactant quantity that you have because
this is what determines the amount of the excess
reactant you will need.
4.13 g Cl2 1 mol Cl2 2 mol NaI 149.89 g NaI 17.5 g NaI
70.90 g Cl2 1 mol Cl2 1 mol NaI
You have 22.1 g available You need 17.5 g 22.1
g NaI 17.5 g NaI 4.6 g NaI will be left over
after the reaction is complete
11 Yield
- Chemists like to know how right we are ?
- yield (actual/theoretical) x 100
- Actual ? lab results
- Theoretical ? mathematical prediction through
stoichiometry - error theoretical experimental/theoretical
x 100 - error yield 100