C3H8(g) + 4 Cl2(g) → C3H4Cl4(g) + 4 HCl(g) 48. A 6.0 mol sample of C3H8(g) and a 20. mol sample of C — Stoichiometry Chemistry Question

STEPS:

1. Write down the balanced chemical equation and given quantities: The balanced equation for the reaction is:
$$\text{C}_3\text{H}_8(g) + 4\ \text{Cl}_2(g) \rightarrow \text{C}_3\text{H}_4\text{Cl}_4(g) + 4\ \text{HCl}(g)$$
The starting quantities placed in the reaction vessel are $6.0\text{ mol}$ of $\text{C}_3\text{H}_8(g)$ and $20.\text{ mol}$ of $\text{Cl}_2(g)$.
2. Understand the concept of the limiting reactant: When reactants are mixed in non-stoichiometric ratios, one reactant will run out first. This substance is the limiting reactant, and it dictates the maximum amount of product that can physically be produced before the reaction stops.
3. Determine the limiting reactant mathematically:
* Calculate how many moles of $\text{Cl}_2$ are required to completely react with all $6.0\text{ mol}$ of $\text{C}_3\text{H}_8$:
$$\text{Moles of }\text{Cl}_2\text{ required} = 6.0\text{ mol }\text{C}_3\text{H}_8 \times \frac{4\text{ mol }\text{Cl}_2}{1\text{ mol }\text{C}_3\text{H}_8} = \mathbf{24\text{ mol }\text{Cl}_2}$$
* Compare this required amount to the actual available amount of $\text{Cl}_2$ ($20.\text{ mol}$). Because the required amount ($24\text{ mol}$) is greater than what is available ($20.\text{ mol}$), we do not have enough chlorine to react with all of the propane.
* Therefore, $\text{Cl}_2$ is the limiting reactant and will be totally consumed. (Propane, $\text{C}_3\text{H}_8$, is in excess).
4. Calculate the theoretical yield of $\text{HCl}(g)$: Using the initial moles of the limiting reactant ($\text{Cl}_2$) and the stoichiometric mole ratio from the balanced equation (4 moles of $\text{HCl}$ produced for every 4 moles of $\text{Cl}_2$ consumed):
$$\text{Moles of }\text{HCl}\text{ produced} = 20.\text{ mol }\text{Cl}_2 \times \frac{4\text{ mol }\text{HCl}}{4\text{ mol }\text{Cl}_2} = \mathbf{20.\text{ mol }\text{HCl}}$$
This matches Option C.

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WHY_OTHERS_WRONG:

• Option A is incorrect (4.0 mol): This value has no correct stoichiometric basis. A student might arrive at this number through a calculation error or by misapplying the coefficients of the products in the chemical equation.
• Option B is incorrect (8.0 mol): This value is obtained if a student incorrectly subtracts the consumed chlorine from the total starting chlorine to find leftover reactant, or performs an incorrect ratio calculation. It does not represent the moles of $\text{HCl}$ product.
• Option D is incorrect (24 mol): This is the number of moles of $\text{HCl}$ that would be produced if propane ($\text{C}_3\text{H}_8$) were the limiting reactant ($6.0\text{ mol }\text{C}_3\text{H}_8 \times 4 = 24\text{ mol }\text{HCl}$). However, because chlorine is the limiting reactant, the reaction runs out of chlorine long before this amount of product can be made.