Stoichiometry Facts

1
Stoichiometry Facts

• Stoichiometry is the relationship between
  reactants and products in a chemical reaction
• Chemical reactions stop when the reactants are
  consumed.
• Equations are balanced to satisfy the Law of
  Conservation of Matter

2
Stoichiometry Facts Contd

• We know from the balanced equation the mole ratio
  between each substance in the equation ( the
  ratio between the coefficients)
• We can use this ratio to mathematically determine
  the amounts of reactants consumed or products
  produced in a chemical rxn.

3
Lets look at an equation
4
So We can..

• Determine the moles, grams, liters, molecules
  (atoms) consumed or produced in a chemical rxn.
• You will be asked to calculate the amount of
  substances related to a given amount of a
  substance.

5
Solving Stoichiometry Problems

• There are 3 basic kinds of stoichiometry
  problems.
• Mass(wt)-Mass(wt),
• Mass-Volume, and
• Vol-Vol.
• The same basic information is needed to work all
  3 types. The only difference is when you see
  volume think 22.4 liters/mole. In these
  problems, the words react with, produce, yield,
  are the key words.

6
Steps to solving Stoichiometry

• Complete the equation( synthesis, single
  replacement, double replacement, decomposition).
• Balance the equation
• Know what the equation tells you (The number of
  moles of each substance reacting and produced).
• Solve by proportion or factor-label

7
How many g of mercury II oxide are needed to
produce 100g of mercury?

• Solution.
• 1. HgO ? Hg O2
• 2 2HgO ? 2 Hg O2
• 3. 2 moles HgO make two moles Hg
• x g 100 g2 HgO ? 2 Hg O2 2(20116)
  2(201) 434 g 402 g

8

• When you use proportion make sure you put your
  given amount and LABEL above the correct
  substance and put X and LABEL about what you are
  asked to find. Only use those substances in
  solving the problem. Use the correct facts to
  find the numbers to go under the specific
  substances. Since this problem asks and uses
  grams we will use the P.T. because we know that 1
  mole of each substance in grams can be found on
  the P.T.Multiply the coefficient (number of
  moles) times the wt off the table and that is the
  number that goes below the substance. Now just
  set up the proportion and solve.

9

• Factor-Label
• 100 g Hg x 1mole Hg x 2 mole HgO x 217g HgO
  201 g Hg 2 mole Hg 1 mole HgO

10

• For factor label always start with your given.
• Now multiply by a fact that will allow us to get
  where we want to go.
• Since we have grams and need moles (why? Because
  that is what the equation tells us. always get
  to moles!!!)
• Use your conversion factor(1 mole mass off P.T.)
  to get from your given to moles.

11

• Next convert from your substance to the
  substance you want by using the coefficients.
  Remember Coeff moles.
• Last convert from moles to the correct label.

12
Another Example

• How many liters of oxygen will be produced when
  150 g of mercury(II) oxide decomposes?This is a
  vol-mass problem

13

• 2 HgO ? 2 Hg O2
• 2 moles HgO make 1 mole O2 150 g
  x liters2 HgO ? 2 Hg O2 2(20116)
  1(22.4) 434 g 22.4 liters
• 150 g x434g 22.4L

14
Factor-Label

• Since this problem uses mass we go to the table
  but the 2nd part uses liters so we now use the
  molar volume information from earlier (1 mole
  22.4 L) NOT PERIODIC TABLE
• 150g HgO x 1mole HgO x 1mole O2 x 22.4liters
  217 g HgO 2 mole HgO 1 mole O2

15

• Exactly the same as before. Using our guide
• we take the given,
• change to moles,
• put coeff of what we want over coeff of what we
  have
• then change to the label we want. Since this
  wants liters we use the fact 1 mole 22.4L

16
Another Example

• How many liters of hydrogen are needed to react
  with nitrogen to produce 400 liters of ammonia
• This is a vol-vol problem

17
Solution.

• 1. N2 H2 ? NH3
• 2. N2 3 H2 ? 2 NH3
• 3. 3 moles H2 make 2 mole NH3
• 4. x liters 400 liters
  N2 3 H2 ? 2 NH3
  3(22.4) 2(22.4) 67.2L
  44.8 liters

18
Factor-Label

• 400 L NH3 x 1 mole NH3 x 3 mole H2 x 22.4 L
  H2 22.4 L NH3 2
  mole NH3 1 mole H2

19
Complications

• The chemical equation is the recipe for the
  reaction. When reactants are added to start a
  chemical rxn, they may be present in any amount.
  Think about our lab. We put in 3g NaHCO3 and
  only 1 drop of HCl. It fizzed and stopped. The
  rxn wasnt complete though because as we added
  more HCl the rxn continued.

20
Contd

• Because we didnt have enough HCl to react with
  all of the NaHCO3. The rxn cant finish unless
  the correct amount of reactants are present.
• The balanced equation tells us the exact amount
  of each substance needed.
• Sometimes we are not given the exact amounts
  necessary.

21
Limiting and Excess

• Reactions proceed until one of the reactants is
  used up and one is left in excess.
• The limiting reactant (reagent) is the substance
  that limits the extent of the reaction and,
  thereby, determines the amount of product formed.
• The excess reactant (reagent) are all the
  leftover unused reactants.

22
Identifying Limiting Reagents

• If we look at the synthesis of water,
• 2H2 O2 ? 2 H2O
• Lets say we are given that we have 20g of
  hydrogen mixed with 40 grams of oxygen. We can
  calculate the moles of each of these.
• 10 mol Hydrogen 1.25 mol oxygen
• Because we have a mole ratio of 21 we know we
  need twice as much Hydrogen as oxygen.

23
Identifying Limiting Reagents Contd

• Our given ratio is 10 mol Hydrogen 1.25 mol
  oxygen or 81
• Since our equation tells us we need a ratio of
  21 we know hydrogen is present in an amount
  greater than we need (excess) and oxygen is
  present in a smaller amount (limiting) so it will
  ultimately determine how much product we can make.

24
Limiting Reactants(Reagents)

• These are special stoichiometry problems. You
  can recognize these stoichiometry problems
  because you will be given of amounts of all the
  reactants and then they will ask for an amount of
  any or all of the products.

25

• The most straightforward way to work this kind is
  to work two problems and the smaller answer is
  the amount of product you can make. The reactant
  that produces that smaller amount is the limiting
  reactant.(Method 1)

26

• Another way is to determine which substance is
  the smaller number of moles.
• If g use P.T. and vol use 22.4. then use the
  ratio of the reactants to find which substance is
  the smaller. Use that one to solve the problem.
  (Method 2)

27
Find the limiting reactant and the amount of
ammonia produced when 6g of hydrogen react with
14g of nitrogen.

• Obviously a limiting reagent problem because it
  asks for it in the question.
• Also you know because you are given the amounts
  of two substances

28
Method 1 Solution.

• 1. N2 H2 ? NH3
• 2. N2 3 H2 ? 2 NH3
• 3. 3 moles H2 make 2 mole NH314g 6g
  x g
• N2 3 H2 ? 2 NH328g 3(2g) 2(17g)14g
  x 17 g ammonia produced by nitrogen28 g
  34g

29
Now do the other part

• 6g x 6 g 34g
• 34 g ammonia produced by hydrogen
• Since 17g is the smaller number, it is the amount
  produced and nitrogen is the limiting reactant

30
Factor-Label

• 14g N2 x 1 mole N2 x 2 mole NH3 x 17g NH3
  28g N2 1 mole N2 1 mole NH3
  17g
• 6g H2 x 1 mole H2 x 2 mole NH3 x 17g NH3
  2g H2 3 mole H2 1 mole NH3
• 34g NH3

31
Method 2 Determine Limiting Rgt

• 14 g nitrogen 0.5 moles and
• 6g hydrogen 3 moles
• the coefficients tell us that we need 3 moles of
  hydrogen to react with 1 mole of nitrogen.
• 0.5 moles N2 need 1.5 moles of H2.
• We have 3 moles H2, which is an excess of what is
  needed. Therefore N2 is the limiting reactant.
• Now work the problem like any mass-mass problem.

32
Yield

• These are normal stoichiometry problems with a
  equation added. In these it might be best to
  remember (actual/theo) times 100.
• The answer to the stoich problem is the amount
  you are supposed to get (theoretical) and the
  amount actually produced is the actual.
• If you like to think part/whole. The calculated
  number is the whole (the total amt you are
  supposed to get) and the amt given as produced is
  the part

33
Calculate the yield if when 14 g of nitrogen
reacts with hydrogen and 15 g of ammonia is
produced

• Solution. 1. N2 H2 ? NH3
• 2. N2 3 H2 ? 2 NH3
• 3. 3 moles H2 make 2 mole NH3
• 4. 14g x g
• N2 3 H2 ? 2 NH3
• 28g 3(2g) 2(17g)

34
Calculate the yield if when 14 g of nitrogen
reacts with hydrogen and 15 g of ammonia is
produced

• Solution.     15g is the actual yield
•      actual x 100
• theoretical
• 15g x 100 88
• 17g

35
Calculate the yield if when 14 g of nitrogen
reacts with hydrogen and 15 g of ammonia is
produced

• Factor-Label
• 14g N2 x 1 mole N2 x 2 mole NH3 x 17g NH3
  28g N2 1 mole N2 1 mole NH3
• 17g
• 15g is the actual yield
•       actual x 100 15 X 100
  88
• theoretical 17

36
Here are your rules

• Factor-Label
• Mass-Mass
• Given(g) x 1 mole x Coeff want x PT g
  PTg Coeff given 1 mole
• Mass-VolGiven(g) x 1 mole x Coeff want x
  22.4 L PTg Coeff
  given 1 mole
• Vol-Vol Given(L) x 1 mole x Coeff want x
  22.4 L 22.4L Coeff
  given 1 mole

37
Here are your rules

• Math (Proportion)
• Mass-Mass
• Given(g) want(g) Coeff x PT
  Coeff x PT
• Mass-Vol Given(g) want(L) Coeff x
  PT Coeff x 22.4L
• Vol-Vol Given(L) want(L) Coeff
  x22.4L Coeff x 22.4L

38
Here are your rules

• Limiting Rgt
• You will have 2 givens.
• Work two problems, use the smallest answer
• Yield
• Work problem
• Take your answer divide it into the amt given

39
Here are your rules-Variations

• Mole-Mole
• Given mol x Coeff want
  Coeff given
• Mole-Mass
• Given(mol) Coeff want x PT g
  Coeff given 1 mole
• What is excess (How much excess)
• Cal amt needed to react with other reactant,
  subtract from the amt given.

40
Here are your rules-Variations

• Mole-Mole
• Given mol want Coeff Coeff
• Mole-Mass
• Given(mol) want
• Coeff Coeff xPT
• What is excess (How much excess)
• Cal amt needed to react with other reactant,
  subtract from the amt given.

41
Calculating Excess Reactant

• When we have a limiting reactant problem we use
  the limiting reagent to calculate the amount of
  product produced.
• We can also use it to determine the amount of the
  excess reactant needed to react with this
  limiting reagent. (Normal Stoich problem)

42
Calculating Amount of Excess Reactant

• Once we calculate the amount of the excess
  reactant that is consumed, we can calculate the
  amount that remains
• (amt given-amt used amt remaining)

43
If we use our example earlier

• 14g 6g x g N2 3 H2 ? 2 NH328g
  3(2g) 2(17g)14g x 3g H2 consumed
  by reacting with 28 g 6g nitrogenNow
  since we started with 6g and used 3g, the amount
  in excess is 6g-3g 3g

44
Factor Label Method

• 14g N2 x 1 mole N2 x 2 mole H2 x 2g H2
  28g N2 1 mole N2 1 mole H2
  3g H2Now take your given and subtract
  the amount used
• 6g-3g 3g