Chemical Equation

1
Chemical Equation

• Describes a chemical reaction
• A B ? C
• A and B reactants
• C product
• CO(g) 2H2(g) ? CH3OH(l)

2
Information Given by Chemical Equations

• CO(g) 2H2(g) ? CH3OH(l)
• 1 molecule of CO 2 molecules of H2 ? one
  molecule of CH3OH
• 10 molecules of CO 20 molecules of H2 ? 10
  molecules of CH3OH
• 1 mole of CO 2 moles of H2 ? 1 mole of CH3OH
• Do you notice a relationship pattern?

3
Stoichiometry

• Process of using a chemical equation to calculate
  the relative masses of reactants and products
  involved in a reaction.

4
Balancing equations

• In a chemical reaction, atoms are neither created
  nor destroyed.
• All atoms present in the reactants must be
  accounted for among the products.
• Therefore, one needs to balance the chemical
  equation for the reaction.
• What goes in must come out, albeit in a different
  form.

5
Rules for balancing equations

• 1. Balance out metals and polyatomic ions first
• Consider polyatomic ions as a single species
• 2. Balance out elements that are found in more
  than one species last
• Leave H for last Na H2O ? NaOH H2
• 3. Dont get discouraged!
• Balancing one atom can throw others off
• Balance the equation afresh
• 4. Do NOT change the nature of the compound
• To balance H2O you cannot make it H2O2
• Why not?
• 5. Thus, put numbers in front of compounds
• 6. Lowest whole numbers
• Incorrect 4A 4B ? 4AB
• Correct A B ? AB

6
Example

• Balance
• Na(s) H2O(l) ? NaOH(aq) H2(g)
• Na is balanced, O is balanced, but H is not
• ? balance H ? put 2 in front of NaOH
• This gives us 4 H on the products side
• Add 2 in front of water on reactants side
• Hs are now balanced
• So are Os (2 on each side)
• Na is now unbalanced
• Add a 2 to Na on reactant side
• Recapitulate check equation
• 2Na(s) 2H2O(l) ? 2NaOH(aq) H2(g)
• BALANCED!

7
Balance the following

• Al(s) F2(g) ? AlF3(s)
• KClO3(s) ? KCl(s) O2(g)
• KI(aq) Pb(NO3)2(aq) ? KNO3(aq) PbI2(s)
• C2H6(g) O2(g) ? CO2(g) H2O(l)
• Lets talk about a neat trick for the above

8
Mole-Mole Relationships Dimensional Analysis

• Consider the decomposition of water
• 2H2O(l) ? 2H2(g) O2(g)
• If we decompose 4 mol of H2O, how many moles of
  O2 do we get?
• If we decompose 5.8 mol of H2O, how many moles of
  H2 do we get?

9
More practice

• Calculate the moles of C3H8 used when 4.30 moles
  of CO2 are obtained.
• C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(g)

10
Mass Calculations

• Steps for Calculating the Masses of Reactants and
  Products in Chemical Reactions
• 1. Balance the equation for the reaction.
• 2. Convert the masses of reactants or products
  to moles.
• 3. Use the balanced equation to set up the
  appropriate mole ratio(s).
• 4. Use the mole ratio(s) to calculate the number
  of moles of the desired reactant or product.
• 5. Convert from moles to mass.

11
In other words

• gramsA ? molesA
• molesA ? molesB
• molesB ? gramsB
• Do it all as a dimensional analysis problem!
• See next slide

12
Example

• What mass of ?2 should we weigh out to react with
  35.0 g Al?
• 2Al(s) 3?2(s) ? 2Al?3(s)
• gramsA ? molesA
• molesA ? molesB
• molesB ? gramsB
• In one-step
• Avoids rounding errors too!

13
Practice

• What mass of CO2 is produced when 96.1 g of C3H8
  react with sufficient O2.
• C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(g)

14
Calculations Involving a Limiting Reactant

• When there is insufficient amount of one of the
  reactants.
• Calculate the mass required for the product in
  question using both reactants.
• Which ever gives the least amount of product is
  the limiting reactant.
• The mass of product obtained from using the
  limiting reactant is the correct amount.

15
Example

• What mass of AlI3 will we obtain if we react 35.0
  g Al with 400.0 g I2?
• 2Al(s) 3I2(s) ? 2AlI3(s)
• 35.0 g Al x (mol/26.98154 g) x (2 mol AlI3/2 mol
  Al) x (407.6950 g/mol) 529 g AlI3
• Versus 400.0 g I2 x (mol/253.8090 g I2) x (2 mol
  AlI3/3 mol I2) x (407.6950 g/mol) 428 g AlI3
• ? I2 is the limiting reactant.
• The mass of AlI3 obtained is 428 g.

16
Practice

• Example
• Calculate the mass of lithium nitride formed from
  56.0 g of N2 and 56.0 g of Li in the balanced
  equation
• 6Li(s) N2(g) ? 2Li3N(s)

17
More practice

• Example
• Calculate the mass of aluminum sulfate formed
  from 50.0 g of Al and 50.0 g of H2SO4 in the
  balanced equation
• 2Al(s) 3H2SO4(aq) ? Al2(SO4)3(s) 3H2(g)

18
Percent Yield

• Theoretical Yield Amount of product predicted
  from the amounts of reactants used.
• Actual Yield Amount of product actually
  obtained. Actual yields are determined by doing
  the reaction
• __Actual Yield__ X 100 Percent Yield
• Theoretical Yield

19
Percent Yield

• Consider the reaction
• TiCl4(g) O2(g) ? TiO2(s) 2Cl2(g)
• (a) Suppose 6.71 x 103 g of TiCl4 is reacted
  with 2.45 x 103 g of O2. Calculate the maximum
  mass of TiO2 that can form.
• (b) If 2.12 x 103 g was produced in the
  laboratory, what is the percent yield?