Stoichiometric Calculation of Maximum Water Produced from Hexane Combustion

Reaction Type: Complete combustion reaction between liquid hexane and gaseous oxygen.
Reactants:
- Liquid hexane ( $C_6H_{14}(l)$ )
- Gaseous oxygen ( $O_2(g)$ )
Products:
- Carbon dioxide ( $CO_2(g)$ )
- Gaseous water ( $H_2O(g)$ )
Given Initial Quantities:
- Mass of hexane ( $C_6H_{14}$ ): $6.03\,g$
- Mass of oxygen gas ( $O_2$ ): $14\,g$
Primary Objective: Calculate the maximum mass of water ( $H_2O$ ) in grams ( $g$ ) that can be produced by the chemical reaction.

Step 1: Balance the chemical equation to ensure that the number of atoms of each element is identical on both the reactant and product sides of the equation. A balanced equation is essential for accurate stoichiometric calculations.
Step 2: Convert grams of hexane ( $C_6H_{14}$ ) to moles of water ( $H_2O$ ).
- Use the molar mass of hexane ( $86.1674\,g\,mol^{-1}$ ) to convert the given mass of hexane to moles of hexane.
- Use the balanced equation's stoichiometric mole ratio to determine the moles of water produced from hexane.
Step 3: Convert grams of oxygen gas ( $O_2$ ) to moles of water ( $H_2O$ ).
- Use the molar mass of oxygen gas ( $31.9988\,g\,mol^{-1}$ ) to convert the given mass of oxygen gas to moles of oxygen gas.
- Use the balanced equation's stoichiometric mole ratio to determine the moles of water produced from oxygen gas.
Step 4: Compare the theoretical moles of water formed from hexane and oxygen gas to determine the limiting reagent. The limiting reagent is the reactant that yields the smallest amount of product.
Step 5: Convert the moles of water generated by the limiting reagent to grams using the molar mass of water ( $18.0151\,g\,mol^{-1}$ ). This yields the maximum mass of water produced.

Initial Unbalanced Equation:
- $C_6H_{14}(l) + O_2(g) \rightarrow CO_2(g) + H_2O(g)$
Sequential Atom Balancing Strategy (Balancing Oxygen Last):
- Carbon Balancing:
- There are $6$ carbon atoms in hexane ( $C_6H_{14}$ ).
- Add a coefficient of $6$ in front of carbon dioxide ( $CO_2$ ) on the product side to balance carbon atoms.
- Intermediate equation: $C_6H_{14}(l) + O_2(g) \rightarrow 6CO_2(g) + H_2O(g)$
- Hydrogen Balancing:
- There are $14$ hydrogen atoms in hexane ( $C_6H_{14}$ ).
- Add a coefficient of $7$ in front of water ( $H_2O$ ) on the product side to balance hydrogen atoms ( $7 \times 2 = 14$ ).
- Intermediate equation: $C_6H_{14}(l) + O_2(g) \rightarrow 6CO_2(g) + 7H_2O(g)$
- Oxygen Balancing:
- Total oxygen atoms on product side: $(6 \times 2) + (7 \times 1) = 12 + 7 = 19$ oxygen atoms.
- To balance oxygen atoms on the reactant side, a coefficient of $9.5$ (or $\frac{19}{2}$ ) is required in front of $O_2$ .
- Intermediate equation with fraction: $C_6H_{14}(l) + 9.5O_2(g) \rightarrow 6CO_2(g) + 7H_2O(g)$
- Eliminating Fractional Coefficients:
- Fractional coefficients must be avoided in standard balanced chemical equations.
- Multiply the entire equation by $2$ to achieve integer coefficients.
Final Balanced Chemical Equation:
- $2C_6H_{14}(l) + 19O_2(g) \rightarrow 12CO_2(g) + 14H_2O(g)$
- Final stoichiometric coefficients:
- $2$ for hexane ( $C_6H_{14}$ )
- $19$ for oxygen gas ( $O_2$ )
- $12$ for carbon dioxide ( $CO_2$ )
- $14$ for water ( $H_2O$ )

Given Quantities and Constants:
- Mass of hexane ( $C_6H_{14}$ ) = $6.03\,g$
- Molar mass of hexane ( $C_6H_{14}$ ) = $86.1674\,g\,mol^{-1}$
- Stoichiometric ratio from balanced equation = $2\,mol\,C_6H_{14} : 14\,mol\,H_2O$
Dimensional Analysis Setup:
- $\text{Moles of } H_2O = 6.03\,g\,C_6H_{14} \times \left(\frac{1\,mol\,C_6H_{14}}{86.1674\,g\,C_6H_{14}}\right) \times \left(\frac{14\,mol\,H_2O}{2\,mol\,C_6H_{14}}\right)$
Unit Cancellation:
- $g\,C_6H_{14}$ cancels with $g\,C_6H_{14}$
- $mol\,C_6H_{14}$ cancels with $mol\,C_6H_{14}$
- Remaining unit = $mol\,H_2O$
Calculated Theoretical Yield from Hexane:
- $\text{Moles of } H_2O = 0.48986\,mol$

Given Quantities and Constants:
- Mass of oxygen gas ( $O_2$ ) = $14\,g$
- Molar mass of oxygen gas ( $O_2$ ) = $31.9988\,g\,mol^{-1}$
- Stoichiometric ratio from balanced equation = $19\,mol\,O_2 : 14\,mol\,H_2O$
Dimensional Analysis Setup:
- $\text{Moles of } H_2O = 14\,g\,O_2 \times \left(\frac{1\,mol\,O_2}{31.9988\,g\,O_2}\right) \times \left(\frac{14\,mol\,H_2O}{19\,mol\,O_2}\right)$
Unit Cancellation:
- $g\,O_2$ cancels with $g\,O_2$
- $mol\,O_2$ cancels with $mol\,O_2$
- Remaining unit = $mol\,H_2O$
Calculated Theoretical Yield from Oxygen Gas:
- $\text{Moles of } H_2O = 0.32238\,mol$

Limiting Reagent Principle: The limiting reagent is defined as the reactant that produces the smallest quantity of product.
Yield Comparison:
- Theoretical yield of $H_2O$ from $6.03\,g$ hexane: $0.48986\,mol$
- Theoretical yield of $H_2O$ from $14\,g$ oxygen gas: $0.32238\,mol$
Evaluation:
- $0.32238\,mol < 0.48986\,mol$
- Oxygen gas ( $O_2$ ) produces fewer moles of water than hexane ( $C_6H_{14}$ ).
Conclusion:
- Limiting Reagent: Oxygen gas ( $O_2$ )
- Excess Reagent: Hexane ( $C_6H_{14}$ )

Maximum Moles of Water Produced:
- Dictated by the limiting reagent ( $O_2$ ), the maximum amount of water that can form is $0.32238\,mol$
Physical Constant:
- Molar mass of water ( $H_2O$ ) = $18.0151\,g\,mol^{-1}$
Conversion Calculation Setup:
- $\text{Maximum Mass of } H_2O = \text{Moles of } H_2O \times \text{Molar Mass of } H_2O$
- $\text{Maximum Mass of } H_2O = 0.32238\,mol\,H_2O \times 18.0151\,g\,mol^{-1}$
Calculated Result:
- $\text{Maximum Mass of } H_2O = 5.8077\,g$

Balanced Reaction Equation: $2C_6H_{14}(l) + 19O_2(g) \rightarrow 12CO_2(g) + 14H_2O(g)$
Starting Mass of Reactants: $6.03\,g$ of hexane ( $C_6H_{14}$ ) and $14\,g$ of oxygen gas ( $O_2$ )
Identified Limiting Reagent: Oxygen gas ( $O_2$ )
Theoretical Molar Yield: $0.32238\,mol\,H_2O$
Maximum Mass of Water Produced: $5.8077\,g$