2. STOICHIOMETRY
1 molecule of O2
2H2 + O2 2H2O
2 molecules of H2 2 molecules of H2O
Now that you know the mole…
1 mole of O2
2H2 + O2 2H2O
2 moles of H2 2 moles of H2O
3. STOICHIOMETRY
1 mole of O2
2H2 + O2 2H2O
2 moles of H2 2 moles of H2O
Think of a reaction as a ratio…
2 moles H2 : 1 mole O2 : 2 moles H2O
This is called a mole ratio
Multiply by 2 = 4 moles H2 : 2 mole O2 : 4 moles H2O
4. STOICHIOMETRY
2 moles H2 : 1 mole O2 : 2 moles H2O
How many molecules of water are formed
when 3.5 moles of O2 react with H2?
= [2 moles H2 : 1 mole O2 : 2 moles H2O] x 3.5
= 7 moles H2 : 3.5 moles O2 : 7 moles H2O
7 moles of water
N = n x NA
= 7 mol water x 6.02x1023 molecules/mol
= 4.21x1024 molecules of water
Therefore 4.21x1024 molecules of water are formed.
5. STOICHIOMETRY
C2H6 + O2 CO2 + H2O
A) How many moles of O2 are required to
react with 13.9mol of C2H6 (ethane)?
B)What volume of H2O would be produced
by 1.40 mol of O2 and sufficient ethane
6. STOICHIOMETRY
C2H6 + O2 CO2 + H2O
Should always balance the equation first
2C2H6 + 7O2 4CO2 + 6H2O
7. STOICHIOMETRY
2C2H6 + 7O2 4CO2 + 6H2O
A) How many moles of O2 are required to react with 13.9mol of C2H6 (ethane)?
iven: mole ratio of C2H6:O2 = 2:7
moles of C2H6 = 13.9mol
2 mol C2H6 = 13.9 mol C2H6
7 mol O2 x
0.2857 = 13.9
x
x = 13.9
0.2857
= 48.6 mol of O2
Therefore 48.6 mol of O2 are required
8. STOICHIOMETRY
2C2H6 + 7O2 4CO2 + 6H2O
What volume of H2O would be produced by 1.40 mol of O2 and sufficient ethane
iven: mole ratio of O2:H2O = 7:6
moles of O2 = 1.40mol
6 mol H2O = x
7 mol O2 1.40 mol O2
0.8571 = x
1.40
x = 0.8571 x 1.40
= 1.20 mol of H2O
But the question asked for volume!!!
9. STOICHIOMETRY
2C2H6 + 7O2 4CO2 + 6H2O
What volume of H2O would be produced by 1.40 mol of O2 and sufficient ethane
x = 1.20 mol of H2O
Given: n = 1.20 mol
MH2O = 18.016 g/mol
m=?
m = nxM
= 1.20 mol x 18.016 g/mol
= 21.6 g
1 gram of water = 1mL at room temperature
Therefore the volume of water produced is 21.6mL
10. STOICHIOMETRY
General rules for solving problems…
STEP 1: Write a balanced chemical equation
STEP 2: If you’re given m or N of a substance, convert it
to the number of moles
STEP 3: Calculate the number of moles of the required
substance based on the number of moles of the given
substance (using the appropriate mole ratio)
STEP 4: Convert the number of moles of the required
substance to mass or number of particles, as directed
by the question
11. STOICHIOMETRY
Passing chlorine gas through molten
sulfur produces liquid disulfur dichloride.
How many molecules of chlorine react to
produce 50.0g of disulfur dichloride?
12. STOICHIOMETRY
Passing chlorine gas through molten sulfur produces liquid
disulfur dichloride. How many molecules of chlorine react to
produce 50.0g of disulfur dichloride?
STEP 1: Write the balanced equation
2S + Cl2 S2Cl2
STEP 2: Convert the given mass of disulfur dichloride
to the number of moles
MS2Cl2 = 135.04g/mol
n = m/M
= 50.0g/135.04g/mol
= 0.37026mol
13. STOICHIOMETRY
Passing chlorine gas through molten sulfur produces liquid
disulfur dichloride. How many molecules of chlorine react to
produce 50.0g of disulfur dichloride?
STEP 3: Calculate the number of moles of the required
substance using your mole ratio
2S + Cl2 S2Cl2
Given: mole ratio of S2Cl2:Cl2 = 1:1
1 mol S2Cl2 = 0.37026 mol S2Cl2
1 mol Cl2 x
x = 0.37026 mol Cl2
14. STOICHIOMETRY
Passing chlorine gas through molten sulfur produces liquid
disulfur dichloride. How many molecules of chlorine react to
produce 50.0g of disulfur dichloride?
STEP 4: Convert the number of moles of chlorine gas to the
number of particles
x = 0.37026 mol Cl2
N = n x NA
= 0.37026mol x 6.02x1023 molecules/mol
= 2.23 x 1023 molecules
Therefore the number of molecules of chlorine is 2.23 x 10 23
16. STOICHIOMETRY
THE LIMITING REACTANT…
2C2H6 + 7O2 4CO2 + 6H2O
Normally, we assume that all reactants are
consumed in a reaction…
Reactants are said to be in stoichiometic
amounts when all reactants are consumed in
the ratios predicted
17. STOICHIOMETRY
THE LIMITING REACTANT…
2C2H6 + 7O2 4CO2 + 6H2O
But there are often reactants that remain
“unreacted”…
O2
O2 O2
O2
O2
GAS
GAS
18. STOICHIOMETRY
THE LIMITING REACTANT…
Lithium nitride reacts with water to
form ammonia and lithium hydroxide.
If 4.87g of lithium nitride reacts with
5.80g of water, find the limiting
reactant.
19. STOICHIOMETRY
THE LIMITING REACTANT…
Lithium nitride reacts with water to
form ammonia and lithium hydroxide.
If 4.87g of lithium nitride reacts with
5.80g of water, find the limiting
reactant.
STEP 1: Write a balanced chemical equation
Li3N + 3H2O NH3 + 3LiOH
20. STOICHIOMETRY
THE LIMITING REACTANT…
Lithium nitride reacts with water to form ammonia and lithium hydroxide.
If 4.87g of lithium nitride reacts with 5.80g of water, find the limiting
reactant.
STEP 1: Write a balanced chemical equation
Li3N + 3H2O NH3 + 3LiOH
STEP 2: Convert the given masses to the number of
moles
nLi3N = m/M
= 4.87g/34.8g/mol
= 0.140mol
nH2O = m/M
= 5.80g/18.0g/mol
= 0.322mol
21. STOICHIOMETRY
THE LIMITING REACTANT…
Lithium nitride reacts with water to form ammonia and lithium hydroxide.
If 4.87g of lithium nitride reacts with 5.80g of water, find the limiting
reactant.
Li3N + 3H2O NH3 + 3LiOH
nLi3N = 0.140mol
nH2O = 0.322mol
STEP 3: Calculate the number of moles of NH3 produced
by both amounts of reactants
1 mol Li3N = 0.140 mol Li3N
1 mol NH3 x
x = 0.140 mol NH3
22. STOICHIOMETRY
THE LIMITING REACTANT…
Lithium nitride reacts with water to form ammonia and lithium hydroxide.
If 4.87g of lithium nitride reacts with 5.80g of water, find the limiting
reactant.
Li3N + 3H2O NH3 + 3LiOH
nLi3N = 0.140mol
nH2O = 0.322mol
STEP 3: Calculate the number of moles of NH3 produced
by both amounts of reactants
3 mol H2O = 0.322 mol H2O
1 mol NH3 x
x = 0.107 mol NH3
5.80g water makes less ammonia than 4.87g lithium nitride
23. STOICHIOMETRY
THE LIMITING REACTANT…
Lithium nitride reacts with water to form ammonia and lithium hydroxide.
If 4.87g of lithium nitride reacts with 5.80g of water, find the limiting
reactant.
Li3N + 3H2O NH3 + 3LiOH
Therefore water is the limiting reactant
24. STOICHIOMETRY
THE LIMITING REACTANT…
P4 + O2 P4O10
A 1.00g piece of phosphorus is burned in a
flask filled with 2.60x1023 molecules of oxygen
gas. What mass of tetraphosphorus
decaoxide is produced?
25. STOICHIOMETRY
THE LIMITING REACTANT…
A 1.00g piece of phosphorus is burned in a flask filled with 2.60x1023 molecules of
oxygen gas. What mass of tetraphosphorus decaoxide is produced?
STEP 1: P4 + 5O2 P4O10
STEP 2: nP = 1.00g P4 no = 2.60x1023 molecules
123.9g/mol P4 6.02x1023molecules/mol
= 8.07x10-3 mol = 0.432 mol O2
LIMITING REACTANT
STEP 3: Amount of P4O10 produced by P4 = 8.07x10-3 mol
Amount of P4O10 produced by O2 = 0.432 ÷ 5
= 8.64 x 10-2 mol
26. STOICHIOMETRY
THE LIMITING REACTANT…
A 1.00g piece of phosphorus is burned in a flask filled with 2.60x1023 molecules of
oxygen gas. What mass of tetraphosphorus decaoxide is produced?
Amount of P4O10 produced by P4 = 8.07x10-3 mol
m= n x M
= 0.00807mol x 284g/mol P4O10
= 2.29g P4O10
Therefore 2.29g of P4O10 is produced.
27. STOICHIOMETRY
THE LIMITING REACTANT…
empty case battery circuitry iphone 4
1 + 1 + 1 1
28. STOICHIOMETRY
THE LIMITING REACTANT…
empty case battery circuitry iphone 4
What if you were given… Limiting ingredient
3 + 6 + 1 1
?
How many iphones can you make?
29. STOICHIOMETRY
THE LIMITING REACTANT…
empty case battery circuitry iphone 4
What if you were given…
101 + 64 + 97 64
?
Limiting ingredient
How many iphones can you make?
