Consider the following intermediate chemical equations:
[tex]\[
\begin{array}{l}
2 P (s) + 3 Cl_2 (g) \rightarrow 2 PCl_3 (l) \\
PCl_3 (l) + Cl_2 (g) \rightarrow PCl_5 (s)
\end{array}
\][/tex]

When you combine the intermediate chemical equations, which substance do you cancel out?

A. \( P \)

B. \( Cl_2 \)

C. \( PCl_3 \)

D. [tex]\( PCl \)[/tex]



Answer :

To find out which substance cancels out when combining the two given intermediate chemical equations, we should follow the steps below:

1. Identify the given equations:

[tex]\[ 2 P (s) + 3 Cl_2 (g) \rightarrow 2 PCl_3 (l) \][/tex]

[tex]\[ PCl_3 (l) + Cl_2 (g) \rightarrow PCl_5 (s) \][/tex]

2. Observe which substances appear in both equations:

- The first equation produces \(2 PCl_3 (l)\).
- The second equation consumes \(PCl_3 (l)\).

3. Combine the two equations:

To combine them, we add the reactions and cancel out any substance that appears on both the reactant and product sides:

[tex]\[ 2 P (s) + 3 Cl_2 (g) \rightarrow 2 PCl_3 (l) \][/tex]

[tex]\[ PCl_3 (l) + Cl_2 (g) \rightarrow PCl_5 (s) \][/tex]

4. Cancel out the common substance:

Since \(PCl_3 (l)\) appears as a product in the first equation and as a reactant in the second equation, it will cancel out when we combine the equations:

[tex]\[ (2 P (s) + 3 Cl_2 (g) \rightarrow 2 PCl_3 (l)) + (PCl_3 (l) + Cl_2 (g) \rightarrow PCl_5 (s)) \][/tex]

Simplifies to:

[tex]\[ 2 P (s) + 3 Cl_2 (g) + PCl_3 (l) + Cl_2 (g) \rightarrow 2 PCl_3 (l) + PCl_5 (s) \][/tex]

After canceling the \(2 PCl_3 (l)\) that appears on both sides:

[tex]\[ 2 P (s) + 4 Cl_2 (g) \rightarrow PCl_5 (s) \][/tex]

Thus, the substance that cancels out when combining the equations is \(PCl_3\).

Therefore, the correct answer is [tex]\(PCl_3\)[/tex].

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