Stoichiometry
STOICHIOMETRY
FOUR STEP SOLUTION
Step 1
Write and balance the reaction.
Step 2
List conversion factors.
Step 3
Set up mole ratios.
Step 4
Solve using dimensional analysis
Stoichiometry: Problem #1
Lithium Nitride reacts with water to form ammonia and
aqueous lithium hydroxide.
Li3N (s) + H2O (l)  NH3 (g) + LiOH (aq)
What mass of water is needed to react with 32.9 g of Li3N ?
1
Step 1: Write and Balance
the equation
Li3N (s) + 3 H2O (l)  NH3 (g) + 3 LiOH (aq)
3
Li
1
1
N
1
3
2
6
H
4
6
1
3
O
1
3
Li3N (s) +
3H2O (l)  NH3 (g) + 3 LiOH (aq)
Step 2:Get conversion factors
Mass/mole conversion factor for Li3N
1 mole of Li3N
= 3 x 6.94 g of Li/mol + 1 x 14.01 g of N/mol
= 34.7 grams Li3N /mol
34.7 g Li3N
1 mol L3N
Mass/mole conversion factor for water (H2O)
1 mole of H2O
= 2 x 1.01 g of H/mol + 1 x 16.00 g of O/mol
= 18.02 grams H2O /mol
18.2 g H2O
1 mol H2O
Step 3: Show the Mole Ratio
Using the balanced equation, write down the
mole ratios for the compounds in the problem
Li3N (s) + 3 H2O (l)  NH3 (g) + 3 LiOH (aq)
problem asked to find grams of water (H2O)
needed to react with 32.9 g of Li3N
The mole ratio is 3 H2O
1 Li3N
2
Li3N (s) +
3H2O (l)  NH3 (g) + 3 LiOH (aq)
find grams of water (H2O) needed
to react with 32.9 g of Li3N
3 H2O
34.7 g Li3N
18.2 g H2O
1 mol L3N
1 mol H2O
1 Li3N
Solve
32.9g Li3N
1mol Li3N
34.7 g Li3N
=
18 g H2O
3 mol H2O
1 mol H2O
1 mol Li3N
51.3 g H2O
Your Turn (and I will help every step of way):


Sample Problem
Tin(II) fluoride, SnF2, is used in some
toothpastes. It is made by the reaction of tin
with hydrogen fluoride according to the
following equation.


Sn(s) + 2HF(g) → SnF2(s) + H2(g)
How many grams of SnF2 are produced
from the reaction of 30.00 g HF with Sn?
Answer Step 1
Tin(II) fluoride, SnF2, is used in some toothpastes. It is made
by the reaction of tin with hydrogen fluoride according to the
following equation:
Sn(s) + 2HF(g) → SnF2(s) + H2(g)
Write and Balance Equations
Sn(s) + 2HF(g) → SnF2(s) + H2(g)
“How many grams of SnF2 are produced from the
reaction of 30.00 g HF with Sn?” Circle what is
important to solve the problem.
3
Step 2
Sn(s) + 2HF(g) → SnF2(s) + H2(g)
2. Write Conversion Factors
molar mass HF molar mass SnF2
mole : mole ratio
2 mol HF
1 mol SnF2
Step 3
Sn(s) + 2HF(g) → SnF2(s) + H2(g)
3. Show Mole Ratio
mole : mole ratio
mole : mole ratio
OR
2 mol HF
1 mol SnF2
1 mol SnF2
2 mol HF
Step 4 Solve the problem
“How many grams of SnF2 are produced from
the reaction of 30.00 g HF with Sn?”
Conversion Factors
molar mass HF molar mass SnF2
mole : mole ratio
2 mol HF
1 mol SnF2
4. Solve
30.00 g HF 1 mol HF
1 mol SnF2 156.71 g SnF2
20.01 g HF 2 mol HF 1 mol SnF2
30 x 156.71 ÷ (20.01 x 2) =
=117.5 g SnF2
117.5
4