Stoichiometry and Limiting Reactants Study Notes

Synthesis and Combination Reactions
- Synthesis reactions, also referred to as combination reactions, involve the combining of reactants to form a single product.
- Example Equation: $4Fe + 3O_2 \rightarrow 2Fe_2O_3$ .
- Checking Balance: An equation must be balanced to ensure the law of conservation of mass is met. In the synthesis of Iron (III) oxide, the balanced coefficients are 4, 3, and 2.
Double Replacement Reactions
- Definition: A reaction where the ions of two compounds exchange places in an aqueous solution to form two new compounds.
- Example: Iron (II) sulfate reacting with Aluminum hydroxide.
- Unbalanced Reaction: $FeSO_4 + Al(OH)_3 \rightarrow Al_2(SO_4)_3 + Fe(OH)_2$
- Balancing Strategy: A useful trick is to start with polyatomic ions. For the right-hand side with three sulfate ions ( $SO_4^{2-}$ ), place a coefficient of 3 before the iron sulfate on the left. To balance hydroxides ( $OH^-$ ), if there are three on the left and two on the right, find a common factor (6), resulting in a 2:3 ratio.
- Balanced Coefficients: 3, 2, 3, 1.

Limiting Reactant
- The limiting reactant is the substance that is totally consumed when the chemical reaction is complete.
- The amount of product formed is limited by this reactant, as the reaction cannot continue once it is used up.
- The reactant that produces the smaller amount of product is definitively the limiting reactant.
Excess Reactant
- The reactant that remains after a limiting reactant is completely consumed.
- Example: In a reaction where iron is limiting and oxygen is excess, excess oxygen will remain after all iron has reacted.

Theoretical Yield
- This is the maximum amount of product that could be formed from the limiting reactant if the reaction went to $100\%$ completion.
Actual (Experimental) Yield
- The amount of product actually produced when the experiment is physically performed in a laboratory setting.
Percent Yield Formula
- Formula: $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$
Interpreting Yield Percentages
- Laboratory Standards: In a reproducible lab experiment, a yield above $70\%$ is generally desired.
- Industrial Standards: In large-scale industrial settings (e.g., ammonia production or pharmaceuticals), a yield of $30\%$ to $40\%$ might be considered acceptable due to the high cost of materials, waste management concerns, and EPA regulations. Producers must balance high yield with low waste and low cost.

Method A: Product Comparison
- Step 1: Start with the mass of each reactant.
- Step 2: Convert the mass of each reactant to moles using their respective molar masses.
- Step 3: Use the mole-to-mole ratio from the balanced equation to find the moles of product each reactant could produce.
- Step 4: Convert the moles of product to grams.
- Step 5: Compare the resulting masses. The reactant yielding the smaller mass of product is the limiting reactant.
Method B: Reactant-to-Reactant Comparison
- Step 1: Convert the given mass/moles of one reactant to the required mass/moles of the second reactant.
- Step 2: Compare the required amount of the second reactant to the actual amount available.
- Step 3: If the amount needed is greater than the amount available, the second reactant is limiting. If the amount needed is less than the amount available, the first reactant is limiting.
Method C: The Molar Quotient Method (The Quickest Method)
- Step 1: Convert all reactant quantities to moles.
- Step 2: Divide the number of moles of each reactant by its stoichiometric coefficient from the balanced equation.
- Step 3: Compare the quotients. The reactant with the smallest quotient is the limiting reactant.
- Note: This method is significantly faster (approx. 3 steps) compared to product comparison (approx. 10 steps).

Scenario: $25\,g$ of Iron reacts with $25\,g$ of Oxygen.
Balanced Equation: $4Fe + 3O_2 \rightarrow 2Fe_2O_3$
Calculations for Iron ( $Fe$ ):
- Molar Mass of $Fe$ : $55.845\,g/mol$
- Moles of $Fe$ : $\frac{25\,g}{55.845\,g/mol}$
- Mole Ratio ( $Fe$ to $Fe_2O_3$ ): 2 to 4.
- Molar Mass of $Fe_2O_3$ : $159.687\,g/mol$
- Calculation: $\frac{25}{55.845} \times \frac{2}{4} \times 159.687 = 35.7\,g\,Fe_2O_3$
Calculations for Oxygen ( $O_2$ ):
- Molar Mass of $O_2$ : $32\,g/mol$ (Diatomic: $16 \times 2$ ).
- Mole Ratio ( $O_2$ to $Fe_2O_3$ ): 2 to 3.
- Calculation: $\frac{25}{32} \times \frac{2}{3} \times 159.687 = 83.2\,g\,Fe_2O_3$
Conclusion: Iron is the limiting reactant because it produces only $35.7\,g$ of product; Oxygen is the excess reactant.
Yield Example:
- If a student produces $20.9\,g$ in lab:
- Percent Yield: $\frac{20.9\,g}{35.7\,g} \times 100 = 58.5\%$

Scenario: $1.5\,mol$ of Methanol ( $CH_3OH$ ) and $0.5\,mol$ of Oxygen ( $O_2$ ).
Equation Observation: Reaction has a $2:3$ ratio for Methanol to Oxygen.
Using Molar Quotient Method:
- Methanol: $\frac{1.5\,mol}{2} = 0.75$
- Oxygen: $\frac{0.5\,mol}{3} = 0.167$
Conclusion: Oxygen is the limiting reactant because $0.167 < 0.75$ .

Reaction: $3FeSO_4 + 2Al(OH)_3 \rightarrow Al_2(SO_4)_3 + 3Fe(OH)_2$
Scenario: $50\,g$ of Iron (II) sulfate ( $FeSO_4$ ) and $50\,g$ of Aluminum hydroxide ( $Al(OH)_3$ ).
Analysis using Method B (Reactant Comparison):
- Molar Mass of $FeSO_4$ : $151.91\,g/mol$
- Molar Mass of $Al(OH)_3$ : $78.004\,g/mol$
- Calculation for needed $Al(OH)_3$ : $\frac{50\,g\,FeSO_4}{151.91\,g/mol} \times \frac{2\,mol\,Al(OH)_3}{3\,mol\,FeSO_4} \times 78.004\,g/mol = 17.1\,g\,Al(OH)_3$
Comparison:
- Amount needed: $17.1\,g$ of $Al(OH)_3$
- Amount available: $50\,g$ of $Al(OH)_3$
- Conclusion: $Al(OH)_3$ is the excess reactant; $FeSO_4$ is the limiting reactant.
Excess Calculation:
- Leftover $Al(OH)_3 = 50\,g - 17.1\,g = 32.9\,g$ .

Question: Is it possible for there to ever be no limiting factor, where both are perfectly consumed?
- Response: While theoretically possible (stoichiometric amounts), in General Chemistry 1 exercises, there will typically be a limiting reactant to identify.
Question: How do you find how much of the excess reactant was used?
- Response: You can back-calculate. Convert the mass of product formed (or the mass of the limiting reactant) back into the units of the excess reactant to see how much was consumed.
Dialogue regarding Lab and Scheduling:
- Student: "Are we doing one trial?"
- Instructor: "We're only doing one trial, but you might have to do several heating and cooling cycles."
- Instructor: "Does other professor waiting outside for something? He's staring at me… He's waving too… Come in… He was just staring at me."