Chemistry Matter and Change: Chapter 11 Stoichiometry

Stoichiometrystudy of quantitative relationships between the amounts of reactants used and the amounts of products formed in a chemical reaction
Mass of Reactants =Mass of Products
Number of atoms of each element on the reactant side of the equation =the number of atoms of each element on the product side of the equation
Number of moles of each element on the reactant side of the equation =the number of moles of each element on the product side of the equation
Mole Ratioratio of the coefficients of any two substances in a balanced chemical equation.
Coefficient of Unknown / Coefficient of Known
4Al + 3O₂→ 2Al₂O₃Mole Ratios:
4 moles Al / 3 moles O₂
or 3 moles O₂ / 4 moles Al
4 moles Al / 2 moles 2Al₂O₃
or 2 moles Al₂O₃/ 4 moles Al
3 moles O₂/ 2 moles 2Al₂O₃
or 2 moles Al₂O₃/ 3 moles O₂
Mole to Mole Conversion (1 step)Given the number of moles of a substance (moles of the Known) in a balanced chemical equation, finding the moles needed of another substance (moles of the Unknown) in the equation.
Moles of Unknown = Moles of known x Mole Ratio
In the equation: 4Al + 3O₂→ 2Al₂O₃,
how many moles of Al₂O₃will be produced if there are 3.75 moles of O₂?
3.75 moles O₂ x 2 moles Al₂O₃/ 3 moles O₂ =
2.50 moles Al₂O₃
In the equation:
C₃H₈ + 5O₂→ 3CO₂+ 4H₂O
How many moles of CO₂are produced when 7 moles of C₃H₈ are burned?
7 moles C₃H₈ x 3 moles CO₂ / 1 mole C₃H₈ =
21 moles CO₂
Mass to Mass Conversion (3 steps)Converting from the mass of the Known to the mass of the Unknown.
Mass of Known / Molar Mass of Known x Mole Ratio x Molar Mass of Unknown
In the equation:
NH₄NO₃ → N₂O + 2H₂O
How many grams of H₂O are produced if you are given 50 g of NH₄NO₃?
50 g NH₄NO₃/ 80.04 g/mol x
2 moles H₂O / 1 mole NH₄NO₃x 18.02 g/mol H₂O =
22.5 g H₂O
Mole to Mass ConversionMoles Known x Mole Ratio x Molar Mass Unknown
In the equation:
2Na + Cl₂→ 2NaCl
How many grams of NaCl are produced from 3.75 moles of Cl₂?
3.75 moles Cl₂ x 2moles NaCl/1 mole Cl₂ x 58.5g/mol NaCl = 438.75 g NaCl
Mass to Mole ConversionMass Known / Molar Mass Known x Mole Ratio
In the equation:
2Na + Cl₂→ 2NaCl
How many moles of Na are needed if 285 g of NaCl are produced?
285g NaCl / 58.5 g/mol x 2 moles Na / 2 moles NaCl =
4.87 moles Na.
How to determine the limiting (and excess) reactants.Conduct two separate mass-to-mass (or mole-to-mole) stoichiometric calculations; comparing the mass of each reactant to a chosen product.
Limiting Reactant:the reactant that limits the amount of product formed in a chemical reaction. It is the reactant that produces the least amount of the chosen product.
Excess Reactant:the reactant that is present in excess; after the reaction has concluded, a portion of this reactant will be left un – reacted. It will be the reactant that produces the greater amount of the chosen product.
Limiting Reactant Problem:Given the following reaction: S₈ + 4Cl₂ → 4S₂Cl₂. If there is 200g of sulfur and 100g of chlorine, what mass of disulfur dichloride will be produced? (This will be determined by which reactant will produce the least amount of disulfur dichloride).
Perform a mass-to-mass calculation between sulfur and disulfur dichloride.200g S₈ x 1mol S₈ / 256.5g S₈ x 4mol S₂Cl₂ / 1 mol S₈ x 135g S₂Cl₂ / 1 mol S₂Cl₂ = 421g S₂Cl₂.
Perform a mass-to-mass calculation between chlorine and disulfur dichloride.100g Cl₂ x 1mol Cl₂ / 70.91g Cl₂ x 4mol S₂Cl₂ / 4 mol Cl₂ x 135g S₂Cl₂ / 1 mol S₂Cl₂ = 190.4g S₂Cl₂.
The limiting reactant is chlorine (Cl₂ ) because…it produced only 190.4g S₂Cl₂.
The excess reactant is sulfur (S₈) because…it would have produced 421g S₂Cl₂.
Determining the amount of excess reactant used and the amount of excess reactant left over in the reaction.Perform a mass-to-mass stoichiometric calculation between the two reactant, using the limiting reactant (Cl₂) as the known quantity and the excess reactant (S₈) as the unknown quantity.
100g Cl₂ x 1mol Cl₂ / 70.91g Cl₂ x 1 mol S₈ / 4mol Cl₂ x 265.5g S₈ = 93.6g S₈Of the 200g of S₈ that was supplied as a reactant, only 93.6g was needed to react with the 100g of Cl₂ (the limiting reactant).
The amount of S₈ in excess is the amount supplied minus the amount used…200g – 93.6g = 106.4g of S₈ in excess
Percent YieldThe ratio of the amount of product that is expected to be produced in a chemical reaction (using stoichiometric calculations based upon the balanced chemical equation) and the amount of product actually produced when carried out in an actual experiment.
Theoretical Yield:the amount of product that is expected to be produced in a chemical reaction (using stoichiometric calculations based upon the balanced chemical equation)
Actual Yield:the amount of product actually produced when an actual experiment is carried out.
Percent Yield Equation:Percent Yield = (Actual Yield ÷ Theoretical Yield) × 100
Percent Yield Problem:Calculate the % Yield of solid silver chromate produced in the following reaction:
K₂CrO₄ + 2AgNO₃ → Ag₂CrO₄ + 2KNO₃
In the reaction there was .500g of the limiting reactant AgNO₃. In the actual experiment, .455g of Ag₂CrO₄ was produced.
Calculating the Theoretical Yield of Ag₂CrO₄ that was produced..500g AgNO₃ x 1 mole AgNO₃ / 169.9g AgNO₃ x
1mole Ag₂CrO₄ / 2 mole AgNO₃ x 331.7 g Ag₂CrO₄ / 1 mol Ag₂CrO₄ = .488g Ag₂CrO₄.
Find the ratio of the actual yield (.455g) to the theoretical yield (.488g) …% Yield = (.455g Ag₂CrO₄ ÷ .488g Ag₂CrO₄ ) x 100 =
93.2 %