1. Chemical Equations and Reaction Stoichiometry

2. The substances that react, called reactants;
CHEMICAL EQUATIONS
Chemical equations are used to describe chemical reactions, and they show
The substances that react, called reactants;
the substances formed, called products; and
(3) the relative amounts of the substances involved.
Number of moles
N.B. The amounts of substances are in moles

3. What does this equation tell us?
it says that for every CH4 molecule that reacts, two molecules of O2 also react, and that one CO2 molecule and two H2O molecules are formed (produced).
If we multiply throughout by Avogadro's number
So, it turns out that
The equation, therefore, tells us that for every CH4 mole that reacts, two moles of O2 also react, and that one CO2 mole and two H2O moles are formed (produced).

4. We know the mass of 1 mol (molar mass) of each of these substances,
Molar mass of CH4 = 12.0 g mol x 1.0 g mol-1 = 16 g mol-1
Molar mass of O2 = 2 x 16.0 g mol-1 = 32.0 g mol-1
Molar mass of CO2 = 12.0 g mol x 16.0 g mol-1 = 44.0 g mol-1
Molar mass of H2O = 2 x 1.0 g mol g mol-1 = 18.0 g mol-1
so we can also write
The equation then tells us that 16.0 g of CH4 reacts with 64.0 g of O2 to produce 44.0 g of CO2 and 36.0 g of H2O

5. In fact knowing the BALANCED chemical equation allows us to know a great deal of information; we can calculate for example;
Number of Moles Formed/Produced
Mass of a Reactant Required
Mass of a Product Formed
The limiting reactant
Percent yield
How?

6. If it is balanced then we can read the equation as
Number of Moles Formed
How many moles of water (H2O) could be produced by the reaction of 3.5 mol of methane (CH4) with excess oxygen (O2) (i.e., more than a sufficient amount of oxygen is present)?
MAKE SURE THAT THE EQUATION IS BALANCED; that is the number of atoms of each element that appears in the LHS appears as well in the RHS.
If it is balanced then we can read the equation as x
3.5 mol
3.5 moles x
Then 7.0 moles of water will be produced

7. Mass of a Reactant Required
What mass of oxygen is required to react completely with 1.20 mol of CH4?
answer
The balanced equation tells us
1.20 moles x
Then 2.40 moles of oxygen are required.
but we are asked about the mass. We therefore need to calculate the mass of 2.40 moles of oxygen.
Mass of one mole of oxygen= molar mass of oxygen (O2)
=16.0 g mol-1x 2= 32.0 g mol-1
The required mass of oxygen (O2)=2.40 mol x 32.0 g mol-1=76.8 g

8. Mass of a Reactant Required
What mass of oxygen is required to react completely with 24.0 g of CH4?
The molar mass of O2= 2 x 16.0 g mol-1=32.0 g mol-1
The molar mass of CH4 = 12.0 g mol x 1.0 g mol-1 = 16.0 g mol-1
The balanced equation tells us
One molar mass of CH4 (MCH4) reacts with molar masses of O2 (MO2)
MCH MO2
24.0 g X
Then 96.0 g of oxygen are required

9. Phosphorus, P4, burns with excess oxygen to form tetraphosphorus decoxide, P4O10. In this reaction, what mass of P4 reacts with 1.50 moles of O2?

10. Mass of a Product Formed
Calculate the mass of CO2, in grams, that can be produced by burning 6.00 mol of CH4 in excess O2.
The molar mass of CO2= 12.0 g mol x 16.0 g mol-1 =44.0 g mol-1
The molar mass of CH4 = 12.0 g mol x 1.0 g mol-1 = 16.0 g mol-1
The balanced equation tells us
One molar mass of CH4 produces ONE molar masses of CO2
6.00 molar mass of CH4 produces X

11. THE CONCEPT OF LIMITING REACTANT
Say we want to me make a sandwich; we then need two slices of bread and one slice of cheese (The sandwich-making equation)
What if we have six slices of bread and two slices of cheese.
Even though you have extra bread, you are limited to making two sandwiches by the amount of cheese you have on hand. Cheese is the limiting reactant.
So the number of sandwiches is determined by the ratio of cheese slices to those of bread

12. When an equal number of molecules of H2 and O2 are reacted according to the equation
2H2+O2 →2H2O
all of the H2 completely reacts, whereas only half of the O2 is consumed. In this case, the H2 is the limiting reactant and the O2 is the excess reactant.

13. What is the limiting reactant in the reaction of g of CH4 with 48.0 g of O2 according to the following equation?
We have 16.0 g of CH4 = 1.0 mol of CH4
We have 48.0 g of O2 = 1.5 mol of O2
Which one is in excess?
Do we have enough CH4 to react with 1.5 mol of O2 ?
The balanced equation tells us
1 mol (CH4) reacts with 2 mol (O2)
X mol (O2)
X=1.5 mol x 1.0 mol / 2.0 mol = 0.75 mol
Then the number of moles of CH4 required to react with 1.5 mol of O2= 0.75 mol
We have 1.0 mol of CH4 i.e. we have excess of CH4
Do we have enough O2 to react with 1.0 mol of CH4 ?
1.0 mol (CH4) X
X=1.0 mol x 2.0 mol / 1.0 mol = 2.0 mol
Then the number of moles of O2 required to react with 1.0 mol of CH4= 2.0 mol
we have only 1.5 mol of O2 i.e. O2 is the limiting reactant
Then we should base any further calculation on the amount of O2

14. What mass of CO2 could be formed by the reaction of 16
What mass of CO2 could be formed by the reaction of 16.0 g of CH4 with 48.0 g of O2?
In here, we are not told, as in previous examples, that one of the reactants exists in excess so we cannot be sure the amount of which reactant should be used in the calculation. So, first we need to know the limiting reactant
We have 16.0 g of CH4 = 1.0 mol of CH4
We have 48.0 g of O2 = 1.5 mol of O2
Which one is in excess?
Do we have enough CH4 to react with 1.5 mol of O2 ?
The balanced equation tells us
1 mol (CH4) reacts with 2 mol (O2)
X mol (O2)
X=1.5 mol x 1.0 mol / 2.0 mol = 0.75 mol
Then the number of moles of CH4 required to react with 1.5 mol of O2= 0.75 mol
We have 1.0 mol of CH4 i.e. we have excess of CH4
Do we have enough O2 to react with 1.0 mol of CH4 ?
1.0 mol (CH4) X
X=1.0 mol x 2.0 mol / 1.0 mol = 2.0 mol
Then the number of moles of O2 required to react with 1.0 mol of CH4= 2.0 mol
we have only 1.5 mol of O2 i.e. O2 is the limiting reactant
Then we should base any further calculation on the amount of O2

15. What mass of CO2 could be formed by the reaction of 16
What mass of CO2 could be formed by the reaction of 16.0 g of CH4 with 48.0 g of O2?
(Continued)
We base our calculation on the amount of O2 because it is the limiting reactant
The balanced equation tells us
2 MO2 produce MCO2
48.0 g X

16. CH4 is therefore the limiting reactant
What mass of CO2 could be formed by the reaction of 16.0 g of CH4 with g of O2?
We have 16.0 g of CH4 = 1.0 mol of CH4
We have g of O2 = 4.0 mol of O2
The balanced equation tells us that one mole of CH4 reacts with two moles of O2.
Then the number of moles of CH4 required to react with 4.0 mol of O2=2.0 mol
we only have 1.0 mol of CH4 i.e. we do not have excess of CH4
Then the number of moles of O2 required to react with 1.0 mol of CH4= 2.0 mol
we have 4.0 mol of O2 i.e. we have excess of O2
CH4 is therefore the limiting reactant
Then we should base our calculation on the amount of CH4
The balanced equation tells us
one molar mass of CH4 produces ONE molar mass of CO2
16.0 g produces X

17. we already have 4.0 mol of CH4
What mass of CO2 could be formed by the reaction of 32.0 g of CH4 with g of O2?
We have 32.0 g of CH4 = 2.0 mol of CH4
We have g of O2 = 4.0 mol of O2
The balanced equation tells us that one mole of CH4 reacts with two moles of O2.
Then the number of moles of CH4 required to react with 4.0 mol of O2= 2.0 mol
we already have 2.0 mol of O2
Then the number of moles of O2 required to react with 2.0 mol of CH4= 4.0 mol
we already have 4.0 mol of CH4
Then the reactants exist in soichiometric ratio and therefore there is no limiting reactant. We can base our calculation on the amount of any one of them.
The balanced equation tells us
one molar mass of CH4 produces ONE molar mass of CO2
32.0 g produces X

18. PERCENT YIELDS FROM CHEMICAL REACTIONS
The theoretical yield from a chemical reaction is the yield calculated by assuming that the reaction goes to completion.
In practice we often do not obtain as much product from a reaction mixture as is theoretically possible. This is true for several reasons.
(1) Many reactions do not go to completion; that is, the reactants are not completely converted to products.
(2) In some cases, a particular set of reactants undergoes two or more reactions simultaneously, forming undesired products (side reactions) as well as desired products.
(3) In some cases, separation of the desired product from the reaction mixture is so difficult that not all of the product formed is successfully isolated.
The term percent yield is used to indicate how much of a desired product is obtained from a reaction.

19. A 15. 6 gram sample of C6H6 was mixed with excess HNO3. We isolated 18
A 15.6 gram sample of C6H6 was mixed with excess HNO3. We isolated 18.0 grams of C6H5NO2. What is the percent yield of C6H5NO2 in this reaction?
answer
First we interpret the balanced chemical equation
We do not need to work out which is the limiting reactant as we are told that we have excess of HNO3
The balanced equation tells us
One molar mass of C6H6 produces ONE molar mass of C6H5NO2
15.6 g produces X

20. CONCENTRATIONS OF SOLUTIONS
The reaction between two solid reactants often proceeds very slowly or not at all. Many chemical reactions are therefore more conveniently carried out with the reactants mixed in solution rather than as pure substances.
A solution is a homogeneous mixture, at the molecular level, of two or more substances. Simple solutions usually consist of one substance, the solute, dissolved in another substance, the solvent.
The solutions used in the laboratory are usually liquids, and the solvent is often water. These are called aqueous solutions.
The general term concentration refers to the quantity of solute in a standard quantity of solution.

21. Molar concentration, or molarity (M), is defined as the number of moles of solute dissolved in one litre (cubic decimeter) of solution.
The unit of molarity is therefore mol/L, mol L-1 or M “read as molar ”
E.g.
An aqueous solution that is 0.15 M NH3 (read this as “0.15 molar NH3”) contains 0.15 mol NH3 per litre of solution.
If you want to prepare a solution that is, for example, M CuSO4, you place mol CuSO4 in a L volumetric flask or place mol CuSO4 in a 0.5 L volumetric flask or …

22. A sample of NaNO3 weighing 0. 38 g is placed in a 50
A sample of NaNO3 weighing 0.38 g is placed in a 50.0 mL volumetric flask. The flask is then filled with water to the mark on the neck, dissolving the solid. What is the molarity of the resulting solution?
To calculate the molarity, we need the number of moles of solute.
Therefore, you first convert grams NaNO3 to moles.
Molar mass of NaNO3= 23.0 g mol g mol-1 + 3x16.0 g mol =84.9 g mol-1
read this as “0.089 molar”

23. The number of moles required is 0.05 mol of CuSO4.5H2O
How many grams of CuSO4.5H2O do we need in order to prepare a 0.2 M solution of CuSO4.5H2O in a 250 mL volumetric flask.
The number of moles required is 0.05 mol of CuSO4.5H2O
But, we are asked about the mass ( how many grams?)
We can calculate the mass of one mole of CuSO4.5H2O i.e. its molar mass then we multiply this molar mass by the number of moles (0.1 mol) to get the required mass
So we need to dissolve g of CuSO4.5H2O in 250 mL water to make 0.2 M solution

24. Let’s prepare a 0.2 M solution of CuSO4.5H2O
mol CuSO4.5H2O (12.48 g)
Water (solvent) is added to bring the solution level to the mark on the neck of the 250-mL volumetric flask.
The molarity of the solution is
mol/0.250 L= M.
The solute is transferred carefully to the volumetric flask

25. Calculate the mass of barium hydroxide Ba(OH)2 required to prepare 2
Calculate the mass of barium hydroxide Ba(OH)2 required to prepare 2.50 L of a 0.06 M solution of barium hydroxide.

26. We need to add to a reaction vessel 0
We need to add to a reaction vessel g of sodium hydroxide, NaOH, in aqueous solution. How many milliliters of M NaOH should be added?

27. DILUTING SOLUTIONS
Number of moles before dilution=Number of moles after dilution=n
Note, in here, M is the molarity of the solution (not the molar mass of the solute)

28. How many milliliters of 18. 0 M H2SO4 are required to prepare 1
How many milliliters of 18.0 M H2SO4 are required to prepare 1.00 L of a 0.900M solution of H2SO4?

29. Arrange the beakers in order of increasing concentration

30. Solution Stoichiometry
If we plan to carry out a reaction in a solution, we must calculate the amounts of solutions that we need. If we know the molarity of a solution, we can calculate the amount of solute contained in a specified volume of that solution.
Calculate the volume in litters of a M solution of sulphuric acid, H2SO4, required to react completely with grams of Na2CO3 according to the equation
The balanced equations tells us
For the reaction to go to completion

31. Find the volume in litres and in millilitres of a 0
Find the volume in litres and in millilitres of a M NaOH solution required to react with 40.0 mL of M H2SO4 solution according to the reaction
The balanced equation tells us that the reaction ratio is 1 mol of H2SO4 to 2 mol of NaOH