solid phosphorus (p) and chlorine (cl₂) gas react to form solid phosphorus pentachloride (pcl₅). suppose you…

Answer:

$\ce{Cl_{2}}$

Explanation:

Step1: Write the balanced equation

The reaction between solid phosphorus ($\ce{P}$) and chlorine gas ($\ce{Cl_{2}}$) to form solid phosphorus pentachloride ($\ce{PCl_{5}}$) is:
$$\ce{P + \frac{5}{2}Cl_{2} -> PCl_{5}}$$
To eliminate fractions, multiply through by 2:
$$\ce{2P + 5Cl_{2} -> 2PCl_{5}}$$

Step2: Determine mole ratios

From the balanced equation, the mole ratio of $\ce{P}$ to $\ce{Cl_{2}}$ is $2:5$ (or $\ce{P}:\ce{Cl_{2}} = 2\ \text{mol} : 5\ \text{mol}$).

Step3: Calculate required moles

We have $11.0\ \text{mol}$ of $\ce{P}$ and $1.0\ \text{mol}$ of $\ce{Cl_{2}}$.

• For $\ce{P}$: If we use all $1.0\ \text{mol}$ of $\ce{Cl_{2}}$, the moles of $\ce{P}$ required are calculated using the ratio:
  $$\text{Moles of } \ce{P} = \frac{2}{5} \times \text{Moles of } \ce{Cl_{2}} = \frac{2}{5} \times 1.0 = 0.4\ \text{mol}$$
  We have $11.0\ \text{mol}$ of $\ce{P}$, which is more than $0.4\ \text{mol}$ needed.
• For $\ce{Cl_{2}}$: If we use all $11.0\ \text{mol}$ of $\ce{P}$, the moles of $\ce{Cl_{2}}$ required are:
  $$\text{Moles of } \ce{Cl_{2}} = \frac{5}{2} \times \text{Moles of } \ce{P} = \frac{5}{2} \times 11.0 = 27.5\ \text{mol}$$
  We only have $1.0\ \text{mol}$ of $\ce{Cl_{2}}$, which is less than $27.5\ \text{mol}$ needed.

Step4: Identify the limiting reactant

A limiting reactant is the one that is completely consumed first. Since we do not have enough $\ce{Cl_{2}}$ to react with all the $\ce{P}$, $\ce{Cl_{2}}$ will be consumed first. Thus, $\ce{Cl_{2}}$ is the limiting reactant.