1 mole = 1 mol = 6.02 x 1023 molecules

Chemical Reaction

Reactants -> Products

aA + bB -> cC + dD

a,b,c,d are coefficients describing the numbers of molecules (moles) reacting together and being formed.

The coefficients can refer to individual molecules or moles of molecules.  If we combine one dozen brown eggs with one dozen white eggs, we will still have a 50:50 ratio of eggs.  We still have one brown egg for every white egg.  It is simply a matter of scale or how many eggs we have.

Chemical reaction equations are balanced

# of atoms of each element in the reactants = # of atoms of each element in the products

# of moles of each element in the reactants = # of moles of each element in the products

Show Fig 6.3

How does this relate to the conservation of mass and atomic theory?

Is the following balanced?

 2H2(g) + O2(g) -> 2 H2O(l) 2 mol H2 + 1 mol O2 -> 2 mol H2O 2 molecules H2 + 1 molecule O2 -> 2 molecules H2O 4 atoms of H + 2 atoms O -> 4 atoms of H 2 atoms of O

Yes, it is balanced!

Is the following balanced?

 CaCl2(aq) + AgNO3(aq) -> AgCl(s) + Ca(NO3)2(aq) 1 atom Ca 2 atoms Cl + 1 atom Ag 1 atom N 3 atoms O -> 1 atom Ag 1 atom Cl + 1 atom Ca 2 atom N 6 atom O

NO!, number of Cl do not balance, # of O do not balance, # of N do not balance.  A balanced version of the reaction is as follows.

 CaCl2(aq) + 2AgNO3(aq) -> 2AgCl(s) + Ca(NO3)2(aq) 1 atom Ca 2 atoms Cl + 2 atom Ag 2 atom N 6 atoms O -> 2 atom Ag 2 atom Cl + 1 atom Ca 2 atom N 6 atom O

DEMO: Show precipitation reaction.

Balancing Chemical Reactions - Trial and Error

Some guidelines:

1. Do not change the subscripts in the reactants and products

 H2(g) + O2(g) -> H2O(l) 2 atom H + 2 atoms O -> 2 atom H 1 atom O

Not balanced.

Don't play with the subscripts.  These identify the chemical substance.  The determine the ration of elements in a particular compound.  Changing them would change the compound!

 H2(g) + O2(g) -> H2O2(l)

NO!  H2O2 is hydrogen peroxide not water!

2. Count the number of each type of atom in the reactants and products

CaC2 + H2O -> Ca(OH)2 + C2H2

Calcium carbide plus water to form calcium hydroxide (slaked lime) + acetylene

Demo: This is the reaction used in carbide miners lamps.

Ca: 1->1
C: 2->2
H: 2->4
O: 1->2   Not balanced

3. Proceed to balance the unbalanced elements by manipulating the coefficients.

In this case need more H and O on reactants side, multiple H2O by 2.

CaC2 + 2H2O -> Ca(OH)2 + C2H2

Ca: 1->1
C: 2->2
H: 4->4
O: 2->2  Balanced!

Many times this requires multiple steps:  For instance, considering the combustion of methane (natural gas)

CH4 + O2 -> CO2 + H2O

C: 1->1
H: 4->2
O: 2->3 start here

CH4 + O2 -> CO2 + H2O

CH4 + 2O2 -> CO2 + H2O

C: 1->1
H: 4->2
O: 4->3 not yet balanced

CH4 + 2O2 -> CO2 + 2H2O

C: 1->1
H: 4->4
O: 4->4  Balanced

4. Finally check and reduce all coefficients to make sure they are the smallest integer ratio.

Could have balanced methane combustion with

2CH4 + 4O2 -> 2CO2 + 4H2O

Divide all by two.

Need to practice these on your own!  As with many things in chemistry, you need to work problems to develop an understanding of the material.

Chemical calculations:

Stoichiometry - the calculation of quantities in a balanced chemical reaction.

Example: Nitrogen fixation (commercially performed by the Haber process)

The coefficients and molecular masses tell us the mass ratios in which compounds combine.

 3H2(g) + N2(g) -> 2NH3(g) 3 mol H2 + 1 mol N2 -> 2 mol NH3 (3)(2)(1.008g) H + (2)(14.0g) N (2)[14.0g + (3)(1.008g)] NH3

a. How many moles of NH3 are produced when 6 moles of H2 react?

H2 = 6 mol
NH3 = ?

Use ratios of coefficients from the balanced chemical equation just as you would conversion factors.  We know for every 3 moles of  H2 we get 2 moles of NH3 and this provides use with two "conversion factors"

(2 moles NH3 / 3 moles H2) or (3 moles H2  / 2 moles NH3).  Just look at "units" to figure out which one to use.

 6 mol H2 2 mol NH3 = 4 mol NH3 3 mol H2

b. How many moles of H2 are needed to react with 14 moles of N2?

 14 mol N2 3 mol H2 = 42 mol H2 1 mol N2

How many moles of N2 are needed to produce 3.4 g NH3?  Now we need to start using molar masses to convert between grams and moles.  To use the coefficients of the chemical reaction to relate amounts of reactants and products, must always use moles, not grams!

i. can convert to moles and rewrite the question:

Molar mass NH3 = 14.01 + 3(1.01) = 17.03 g / mol

 3.4 g NH3 1 mol NH3 = 0.20 mol H2 17.03 g NH3

Rewritten: How many moles of N2 are needed to produce 0.20 mol NH3

 0.2 mol NH3 1 mol N2 = 0.10 mol N2 2 mol NH3

c. What if the question asked how many grams of N2 are needed to produce 3.4 g NH3?

Would do the same thing as above + one final calculation using the molar mass of N2= 2(14.01) = 28.02 g / mol

 0.10 mol N2 28.02 g = 2.8 g N2 1 mol N2

Can also do this as one big calculation:

How many grams of N2 are need to produce 3.4 g NH3?

 3.4 g NH3 1 mol NH3 1 mol N2 28.02 g = 2.8 g N2 17.03 g NH3 2 mol NH3 1 mol N2 start with grams convert to moles use reaction stoichiometry convert back to grams