Stoichiometry Review
Mole Ratios
The coefficients in a chemical equation represent the relative numbers of moles of reactants and products.
The ratio of the coefficient of one molecule or formula unit in a balanced chemical equation to the coefficient of another one
in the equation.
Therefore:
A balanced chemical equation allows one to determine the mole ratio of any two substances in the reaction
2KClO3 →
2KCl + 3O2
In the reaction above the mole ratio of potassium chlorate to potassium chloride is 1:1
In the reaction above the mole ratio of potassium chlorate to oxygen is 2:3
CH4 + 2O2 → CO2 + 2H2O
In the reaction above the mole ratio of methane to water is 1:2
In the reaction above the mole ratio of methane to carbon dioxide is 1:1
Stoichiometry (You need a balanced equation First)
These can be used to solve stoichiometry problems: mass, molar mass, mole ratios, other conversion factors.
SF4 + 2H2O → SO2 + 4HF
For the reaction above how much water is needed to react 22.0 grams sulfur tetrafluoride?
You need to calculate the molar masses of sulfur tetrafluoride and water first.
SF4
S 1 × 32.06 = 32.06
F 4 × 19.00 = 76.00
108.06
H2O
For the reaction above how much hydrogen monofluoride could be produced with 10.0 grams of
water? You need the molar mass of HF now.
H 2 × 1.01 = 2.02
O 1 × 16.00 = 16.00
18.02
HF
H 1 × 1.01 = 1.01
O 1 × 16.00 = 16.00
17.01
Limiting/Excess Reactants
The limiting reactant is the least abundant and most expensive. The excess reactant is the most abundant and least expensive.
When substance A and substance B are reacted together and substance B is completely used, B is the limiting reactant: there are
insufficient amounts of B to react with all of substance A. Substance A must then be the excess reactant. Some A will be left
over: A is not completely used up.
To determine which reactant is limiting, one must know the available amount of each reactant.
Steps:
Calculate the amount of one of the products with the masses of the two reactants.
Cl2(g) + 2LiI(s)
→ 2LiCl(s) + I2(s)
If 51 grams of chlorine gas and 61 grams of solid lithium iodide react, how much lithium chloride and how much iodine can be made.
You would have to calculate the molar masses of each reactant and product. We’ll assume you know how to do that by now. Be
careful! Chlorine has two atoms and Iodine has two atoms each molecule. (Cl2: 70.90 g/mol; LiI: 133.85 g/mol; LiCl: 42.39 g/mol;
I2: 253.81 g/mol)
1.
Run the mass of chlorine first and pick one of the products. Since both products are asked for let’s pick the first one, LiCl.
2.
Run the mass of lithium iodine now and calculate the same product as in step 1.
3.
Which one limits the amount of product? Which one produces the lesser amount of product? It is LiI even though more mass
of LiI was used than the mass of Cl2. LiI is the limiting reactant.
4.
The amount of LiCl that can be made then is 19g.
5.
What amount of I2 can be made? Use the limiting reactant mass which is 61 g LiCl.
If the reaction problem states that one of the reactants is in excess, then the other reactant must be the limiting reactant.
Yield
Theoretical Yield is the maximum amount of product calculated that can be produced based on the Stoichiometry.
Actual Yield
o Must be determined by experiment
o Is usually lower than the theoretical yield. Otherwise the Law of conservation of mass would be violated.
Percentage Yield
o
When 61 grams of solid lithium iodide reacted with excess chlorine gas, 50.0 grams of iodine was produced. Calculate the percentage
yield of iodine.
Cl2(g) + 2LiI(s) → 2LiCl(s) + I 2(s)
1.
What amount of I2 can be Theoretically made? Use the limiting reactant mass which is 61 g LiCl to calculate theoretical.
2.
Calculate the percentage yield. 59 g is the theoretical.
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