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Chemical Quantities
Chapter 9
9.2 Mole-Mole Relationships
2H2O(l) → 2H2(g) + O2(g)
2 moles of water breaks down into 2 moles
of Hydrogen and 1 mole of Oxygen
How many moles of products are formed
when we break down 4 moles of water?
Multiply everything by 2
2[2H2O(l) → 2H2(g) + O2(g)]
4H2O(l) → 4H2(g) + 2O2(g)
What about a number like 5.8 moles?
Make it easier by dividing everything by 2
so we can start with 1 mole of water
[2H2O(l) → 2H2(g) + O2(g)]/2
H2O(l) → H2(g) + ½ O2(g)
Now simply multiply everything by 5.8
5.8[H2O(l) → H2(g) + ½ O2(g)]
5.8H2O(l) → 5.8H2(g) + 2.9O2(g)
What if we want to find moles required when
there are 2 reactants?
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Calculate the number of moles of oxygen
required to react exactly with 4.3 moles of
propane, C3H8, in the above reaction
4.3 moles of C3H8 requires how many moles of
O2
There is a 1:5 ratio
So 4.3(1) : 4.3(5)
4.3 = 21.5
4.3 mol C3H8(g) + 21.5 O2(g)
How many moles of products are formed?
9.3 Mass Calculations
We can do the same sort of thing with the
mass of the elements and compounds
Let’s use same chemical equation
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Instead of moles we will use mass which is
more realistic
If we have 44.1 g of propane, how much
oxygen will it react with?
Let’s use what we know again
We can convert grams of propane to
moles of propane
1 mole of propane is 44.09 g of propane
So 44.1 g x 1 mole / 44.09 g = 1 mole
Know we need 5 moles of oxygen for
every mole of propane
1 mol C3H8 x 5 mol O2 / 1 mol C3H8 = 5 mol O2
Now how many grams of O2 are in 5 mol
5 mol x 32 g / 1 mol = 160 g
9.5 Mass Calculations: Comparing
Two Reactions
NaHCO3(s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)

Reaction of baking soda and stomach acid
Mg(OH)2(s) + 2HCl(aq) → 2H2O(l) + MgCl2(aq)

Reaction of Milk of Magnesia and stomach acid
Looking at these 2 reactions, which antacid can
consume the most stomach acid, 1.0 g of
NaHCO3 or 1.0 g of Mg(OH)2(s)?
What are we trying
to figure out?
Want to know which
reacts with the most
moles of HCl
So need to do a mole
to mole ratio
First must change
grams of each antacid
to moles
Find the number of moles is in each mass


The molar mass of NaHCO3 is 84.01g so 1.00 g
NaHCO3 x 1 mole / 84.01 g = 0.0119 mol
The molar mass of Mg(OH)2 is 58.33 g so 1.00 g
Mg(OH)2 x 1 mole / 58.33 g = 0.0171 mol
Now we use the mole ratio for each


1 mole NaHCO3 / 1 mole HCl = 0.0119 mole / x
1 mole Mg(OH)2 / 2 mole HCl = 0.0171 mole / x
When we do the math we see that


0.0119 mole NaHCO3 reacts with 0.0119 mole HCl
0.0171 mole Mg(OH)2 reacts with 0.0342 mole HCl
So which 1 gram sample of antacid
consumes the most stomach acid?
A 1 g sample of Milk of Magnesia will consume
more acid
9.6 The Concept of Limiting
Reactions
When you have
multiple ingredients
involved in producing
something, one will
run out first and
cause a limit on how
much product you can
make
Remember the
Sandwich making
example!!
CH4(g) + H2O(g) → 3H2(g) + CO(g)
What mass of H2O(g) is needed for 249 g of
CH4(g)? (same steps we’ve been using)




Need to make sure equation is balanced
Need to convert grams CH4(g) to moles CH4(g)
Need to compare moles CH4(g) to moles H2O(g)
Need to change moles H2O(g) to grams H2O(g)
Let’s do the Problem
Is Equation Balanced? Yeah!!
Convert 249 g CH4 to moles CH4

249g x 1mole/16g = 15.5 mol CH4
Change moles CH4 to moles H2O


15.5 mol CH4 x 1 mol H2O / 1 mol CH4 = 15.5 mol H2O
As proportion 1 mol CH4/15.5 mol CH4 = 1 mol H2O/ ? mol H2O
Change moles H2O to grams H2O
 15.5 mol H2O x 18 g / 1 mol = 279 g H2O
So in order for us to use up 249 g methane we need to
have 279 g water
But what if we didn’t have the right
amount of each?
Well, 1 of them would be a limiting reactant
To solve we will use the same steps as before
with a little twist
Steps for Solving Limiting Reactant
Problems
Write and Balance Equation
Convert Masses of Reactants to
Moles
Use Mole Ratios to Determine what is
Limiting
Use Amount of Limiting Reactant and
Mole Ratios to Compute the Number
of Moles of Product
Convert Moles to Grams (if needed)
Problem
25,000 g Nitrogen gas and 5,000 g Hydrogen
gas react to form ammonia. How much
ammonia is made when reaction stops?
Step 1

N2 + 3H2 → 2NH3
Step 2


25,000 g N2 x 1 mole N2 / 28 g N2 = 892 mol N2
5,000 g H2 x 1 mole H2/ 2.0 g H2 = 2,500 mol H2
Step 3 (just pick one)

892 mol N2 x 3 mol H2 / 1 mol N2 = 2,680 mol H2
Or

2,500 mol H2 x 1 mol N2 / 3 mol H2 = 833.33 mol N2
This tells us we need 2,680 mol H2 and 833.33
mol N2 to use up all the reactants, do we have
these amounts?

No, we do not have enough H2 (only have 2,500
mol and need 2,680 mol) so it will run out = limiting
reactant
Step 4

2,500 mol H2 x 2 mol NH3/3mol H2 = 1,650 mol NH3
Step 5

1,650 mol NH3 x 17 g NH3/ 1 mol NH3= 28,100 g NH3
Practice !!
9.8 Percent Yield
Products stop forming when 1 reactant
runs out
Theoretical Yield = how much you would
get in a perfect world
Actual Yield = what you actually get
Percent Yield = comparing these two

Actual / theoretical x 100 = % yield