Transcript Limiting Reactant and Percent Yield

Limiting Reactant
and Percent Yield
The reactant that is consumed
first and limits the formation of
products
Limiting Reactant

When 2 substances are allowed to react one of them usually reacts
completely while the other is not all used up because it is present in
excess

The reactant that is completely used up in the reaction is the
limiting reactant. It determines the amount of products formed. The
other reactant that is not completely consumed in the reaction is the
excess reactant
Percent Yield
In chemical calculations, it is always assumed that the
reaction goes to completion. However in reality, this
seldom happens. The expected amount of product is not
usually obtained
percent yield = actual yield x 100%
theoretical yield
Actual yield -> amount of product formed from the actual
chemical reaction and is usually less than the theoretical
yield
Theoretical yield  the maximum amount of products
which could be produced by the complete reaction of the
limiting reactant
Example1
2Al + 3I2 ----> 2AlI3
Determine the limiting reactant if one starts with
a)
1.20mol Al and 2.40 mol I2
b)
1.20gAl and 2.40g I2
Solution
a)
Since the given is already in moles, we use those numbers directly. To
find the limiting reactant: take the moles of each substance and divide it
by the coefficient of the balanced equation.
Al : 1.20/2 = 0.6
I2 : 2.40/3 = 0.80
the lowest number indicates the limiting reactant
b)
Since we have grams, we must first convert to moles. Then multiply it by
the molar ratio of the reactant to product.

1.20g Al x 1molAl = 0.044molAl x 2molAlI3 = 0.044 AlI3
26.98gAl
2molAl
2.40gI2 x 1molI2= 0.0095molI2 x 2molAlI2 = 0.0033 molAlI3
253.8g I2
3mol I2
I2 is the limiting reactant since it produces less amount of the product
Example
In an experiment, 5.00g aluminum is heated with 25gS to form aluminum sulfide.
The equation for the reaction is
2Al(s) + 3S(s) ------> Al2S3(s)
a)
How many grams of aluminum sulfide will be formed?
b)
How many grams of excess reactant will remain.
Solution
Step1: Identify the limiting reactant by calculating the amount of product formed
from each of the given amounts of reactants.
5g Al x 1mol Al = 0.185 mol Al
26.98gAl
From the balanced equation, 2 mol Al forms 1 molAl2S3, therefore
mole Al2S3: 0.185mol Al x 1mol Al2S3 = 0.0925mol Al2S3
2mol Al
25gS x 1mol S = 0.780molS
32.07gS
From the balance equation 3moles S forms 1mole Al2S3 therefore
Mole Al2S3 : 0.780S x 1molAl2S3 = 0.260mol Al2S3
3molS
5gAl produces less amount of Al2S3 therefore, it is the limiting reactant
Calculate the mass of Al2S3
0.0925Al2S3mol x 150.17g Al2S3 = 13.9gAl2S3
1molAl2S3
b. Find the excess amount by determining the actual amount of S that reacted
with the limiting reactant, Al
From the balanced equation, 2 moles Al reacts with 3 moles S; therefore
mole S : 0.185mol Al x 3mol S = 0.278mol S
2mol Al
Since the available S is 0.780mol then the amount of S that has not reacted is
0.780mol-0.278mol = 0.502mol S
Calculating the mass excess
0.502mol S X 32.07gS = 16.1g
1mol S
Sample Problem
Wine is produced by the fermentation of fruit sugar, fructose, to alcohol. The
chemical reaction is
C6H12O6 ---> 2C2H6O + 2CO2
If 938g of fructose was used in the preparation of wine, what is the percent yield if, after
the fermentation,327g ethanol was produced?
Find: % yield
Solution: % yield = actual Yield
x 100%
theoretical yield
1.
Determine the theoretical yield
Convert mass C6H12O6 to moles using molar mass and calculate the maximum number
of moles of C2H6O which could be produced using the molar ratio in the balanced
equation
938gC6H12O6 x 1mole C6H12O6 x 2molC2H6O = 10.4molC2H6O
180gC6H12O6
1molC6H12O6
then convert 10.4 molC2H6O using its molar mass (46.07g/mol)
10.4mol C2H6O x 46.07g
= 479g
1mol C2H6O
2. Calculate the percent yield = Act. Yield x 100% = 327g x100% = 68.3%
Theo.Yield
479g

Practice Problems
1.Consider the reaction: Se + 3BrF5 ----> SeF6 + 3 BrF3 , if 0.270mole Se
reacts with BrF and 83.4gBrF3 is formed, what is the percent yield of the
reaction?
2.The reaction: 4NH3 +5 O2 ---> 4NO + 6H2O is one of the steps in the
commercial processes for converting ammonia(NH3) to nitric acid(HNO3).
In a certain experiment, 4.50gNH3 is reacted with 7.50gO2
a) Which is the limiting reactant?
b) How many grams of H2O are formed assuming 100%yield?
c) How many grams of excess reactant remain after the reaction?
Answers:
1.
% yield = 75.1%
2. a) O2 b) 5.06gH2O
c) 1.41gNH3