Unit 9 Stoichiometry

1
Unit 9 Stoichiometry

• Mole-Mole, Mass-Mass, Mass-Volume, Volume-Volume,
  Limiting Reactant and Percent Yield

2
Stoichiometry

• The study of quantitative, or measurable
  relationships that exist in chemical formulas and
  chemical reactions.
• One form of stoich involves converting among
  moles, particles, mass and volume of chemical
  formulas. (Unit 8)
• The other form of stoich is concerned with
  chemical reactions and involves the relationships
  between reactants and products in a chemical
  reaction. (Unit 9)

3
Mole-Mole Problems

• The coefficients in a balanced chemical equation
  can be interpreted both as the number of
  particles and as the number of moles involved in
  the reaction.
• Observe the chemical equation below.
• 2H2 O2 ? 2H2O
• According to the equation 2 moles of hydrogen
  and one mole of oxygen produce 2 moles of water.

4
Mole-Mole Problems

• Suppose you wanted to know how many moles of
  hydrogen and oxygen are needed to produce 5.48
  moles of water.
• First determine the molar ratios indicated by the
  equation.
• 2H2 O2 ? 2H2O
• H2 to H2O is 22
• O2 to H2O is 12

5
Mole-Mole Problems

• Second, set up equations to convert the number of
  moles of the reactant to the number of moles of
  each of the products using dimensional analysis.
• 5.48 mol H2O 2 mol H2 5.48 mol H2
• 2 mol H2O
• 5.48 mol H2O 1 mol O2 2.74 mol O2
• 2 mol H2O

6
Mole-Mole Problems

• How many moles of FeCl3 will be produced from 5
  moles of Cl2?
• 2FeBr3 3Cl2 ? 2FeCl3 3Br2
• 5 mol Cl2 2 mol FeCl3 3.33 mol FeCl3
• 3 mol Cl2

7
Mass-Mass Problems

• You will be given the mass of one substance and
  are asked to find the mass of another substance
  involved in the same chemical reaction.

8
Mass-Mass Problems

• Determine the mass of NaOH produced when 0.25g of
  Na reacts with water according to the following
  equation
• 2Na 2H2O ? 2NaOH H2
• 0.25g Na 1 mol Na 2 mol NaOH 39.99711g
  NaOH
• 22.98977gNa 2 mol Na 1
  mol NaOH
• 0.43g NaOH

9
Mass-Mass Problems

• How many grams of aluminum chloride will be
  produced from 92.00g of Cl2?
• 2AlBr3 3Cl2 ? 3Br2 2AlCl3
• 92.00g Cl2 1 mol Cl2 2 mol AlCl3 133.341g
  AlCl3
• 70.906g Cl2 3 mol Cl2
  1 mol AlCl3
• 115.3g AlCl3

10
Mass-Volume Problems

• Used to calculate the volume of gas produced when
  the mass of another substance in the reaction is
  known.

11
Mass-Volume Problems

• How many liters of oxygen are necessary for the
  combustion of 340.00g of ethanol (C2H5OH)
  assuming that the reaction occurs at STP?
• C2H5OH 3O2 ? 2CO2 3H2O
• 340.00g C2H5OH 1 mol C2H5OH 3 mol O2
  22.4 L O2
• 46.069g C2H5OH 1 mol C2H5OH 1 mol O2
• 495.95 L O2

12
Mass-Volume Problems

• How much hydrogen gas, at STP, is formed from the
  reaction of 3.50g of zinc with an excess amount
  of hydrochloric acid?
• Zn 2HCl ? ZnCl2 H2
• 3.50 g Zn 1 mol Zn 1 mol H2 22.4 L H2
  
• 65.39g Zn 1 mol Zn 1 mol H2
• 1.20 L H2

13
Volume-Volume Problems

• The coefficients in a chemical equation represent
  the ratio of the volumes of gases involved in the
  reaction.
• Very similar to Mole-Mole problems.

14
Volume-Volume Problems

• What volumes of hydrogen and nitrogen gases are
  necessary to produce 16.0 L of ammonia gas?
• N2 3H2 ? 2NH3
• 16.0 L NH3 1 mol NH3 1 mol N2 22.4 L N2
  8.00 L N2
• 22.4 L NH3 2 mol NH3 1
  mol N2
• 16.0 L NH3 1 mol NH3 3 mol H2 22.4 L H2
  24.0 L H2
• 22.4 L NH3 2 mol NH3 1 mol
  H2

15
Limiting Reactants

• The amount of product that can be produced from a
  chemical reaction depends on the amounts of
  reactants available.
• Stoichiometric proportion when the quantities
  of reactants are available in the exact ratio
  described by the balanced equation.

16
Limiting Reactants

• There is usually more of one reactant than can be
  used this reaction is said to be in
  nonstoichiometric proportions.
• The reactant that gets used up first and
  therefore limits the amount of product formed is
  called the limiting reactant.

17
Limiting Reactants

• To solve a limiting reactant problem, you must
  complete two separate mass-mass problems. First
  calculate the mass of product formed using the
  given mass of one reactant. Then calculate the
  mass of product formed using the given mass of
  the other reactant.
• The limiting reactant is the one that produces
  the smaller amount of product.

18
Limiting Reactant Problems

• Identify the limiting reactant when 10.0 g H2O
  reacts with 3.5g Na to produce NaOH and H2.
• 2Na 2H2O ? 2NaOH H2
• 10.0 g H2O 1 mol H2O 2 mol NaOH 39.9971 g
  NaOH
• 18.0153g H2O 2 mol H2O 1 mol NaOH
• 22.2 g NaOH

19
Limiting Reactant Problems

• Identify the limiting reactant when 10.0 g H2O
  reacts with 3.5g Na to produce NaOH and H2.
• 2Na 2H2O ? 2NaOH H2
• 3.5g Na 1 mol Na 2 mol NaOH 39.9971 g
  NaOH
• 22.98977g Na 2 mol Na 1 mol NaOH
• 6.1 g NaOH

20
Limiting Reactant Problems

• NaCl reacts with Pb(NO3)2 to produce PbCl2 and
  NaNO3. What is the limiting reactant when 45.3g
  of NaCl reacts with 60.8g of Pb(NO3)2?
• Pb(NO3)2 2NaCl ? PbCl2 2NaNO3
• 60.8g Pb(NO3)2 1 mol Pb(NO3)2 1 mol PbCl2
  278.1054g PbCl2
• 331.2098g Pb(NO3)2
  1mol Pb(NO3)2 1 mol PbCl2
• 51.1g PbCl2

21
Limiting Reactant Problems

• NaCl reacts with Pb(NO3)2 to produce PbCl2 and
  NaNO3. What is the limiting reactant when 45.3g
  of NaCl reacts with 60.8g of Pb(NO3)2?
• Pb(NO3)2 2NaCl ? PbCl2 2NaNO3
• 45.3g NaCl 1 mol NaCl 1 mol PbCl2
  278.1054g PbCl2
• 58.44247g NaCl 2 mol NaCl
  1 mol PbCl2
• 108g PbCl2

22
Limiting Reactant Problems

• Identify the limiting reactant when 12.5 L H2S at
  STP is bubbled through a solution containing
  24.0g KOH to form K2S and H2O.
• H2S 2KOH ? K2S 2H2O
• 12.5 L H2S 1 mol H2S 2 mol H2O
  18.01528g H2O 20.1g H2O
• 22.4L H2S 1 mol H2S
  1 mol H2O
• 24.0g KOH 1 mol KOH 2mol H2O
  18.01528g H2O 7.71gH2O
• 56.10564g KOH 2mol KOH
  1 mole H2O

23
Percent Yield

• The percent of the expected yield which was
  actually obtained experimentally.
• Percent Yield actual yield x 100
• theoretical yield
• Actual Yield- the amount of a product that is
  really obtained from a chemical reaction.
• Theoretical Yield- the amount of product that
  should be produced based on calculations.

24
Percent Yield

• Determine the percent yield for the reaction
  between 2.80g Al(NO3)3 and excess NaOH if 0.966g
  of Al(OH)3 is recovered.
• First find Theoretical Yield
• Al(NO3)3 3NaOH ? Al(OH)3 3NaNO3
• 2.80gAl(NO3)3 1 mol Al(NO3)3 1mol Al(OH)3
  78.00356gAl(OH)3
• 212.996gAl(NO3)3 1mol
  Al(NO3)3 1mol Al(OH)3
• 1.03g Al(OH)3

25
Percent Yield

• Actual Yield 0.966g Al(OH)3
• Theoretical Yield 1.03g Al(OH)3
• Yield Actual Yield x 100
• Theoretical Yield
• Yield 0.966g x 100 93.8
• 1.03g