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To determine which substance gets canceled out when combining the given intermediate chemical equations, let's look at each reaction and combine them step-by-step.
Consider the intermediate chemical equations provided:
[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]
### Step-by-Step Combination
1. Write down both equations:
First equation:
[tex]\[ 2 P (s) + 3 Cl_2 (g) \rightarrow 2 PCl_3 (l) \][/tex]
Second equation:
[tex]\[ PCl_3 (l) + Cl_2 (g) \rightarrow PCl_5 (s) \][/tex]
2. Search for the common substance:
Identify a substance that appears on both sides of the reactions. Specifically, we see that [tex]\(PCl_3\)[/tex] appears as a product in the first reaction and as a reactant in the second reaction.
3. Combine the equations:
Combine the two equations and make sure to cancel out any substances that appear on both sides:
From the first equation:
[tex]\[ 2 P (s) + 3 Cl_2 (g) \rightarrow 2 PCl_3 (l) \][/tex]
From the second equation:
[tex]\[ PCl_3 (l) + Cl_2 (g) \rightarrow PCl_5 (s) \][/tex]
In combining these, the substance [tex]\(PCl_3 (l)\)[/tex] appears on the product side in the first equation and on the reactant side in the second equation. Thus, we can cancel out [tex]\(PCl_3 (l)\)[/tex]:
[tex]\[ 2 P (s) + 3 Cl_2 (g) \rightarrow 2 PCl_3 (l) \][/tex]
[tex]\[ 2 PCl_3 (l) + 2 Cl_2 (g) \rightarrow 2 PCl_5 (s) \][/tex]
Simplifying:
[tex]\[ 2 P (s) + 3 Cl_2 (g) \rightarrow 2 PCl_3 (l) \rightarrow 2 PCl_3 (l) + 2 Cl_2 (g) \rightarrow 2 PCl_5 (s) \][/tex]
After canceling [tex]\(PCl_3 (l)\)[/tex], we get:
[tex]\[ 2 P (s) + 3 Cl_2 (g) + 2 Cl_2 (g) \rightarrow 2 PCl_5 (s) \][/tex]
Simplify further:
[tex]\[ 2 P (s) + 5 Cl_2 (g) \rightarrow 2 PCl_5 (s) \][/tex]
### Conclusion
When combining the intermediate chemical equations, the substance that gets canceled out in the overall reaction is [tex]\( PCl_3 \)[/tex].
Consider the intermediate chemical equations provided:
[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]
### Step-by-Step Combination
1. Write down both equations:
First equation:
[tex]\[ 2 P (s) + 3 Cl_2 (g) \rightarrow 2 PCl_3 (l) \][/tex]
Second equation:
[tex]\[ PCl_3 (l) + Cl_2 (g) \rightarrow PCl_5 (s) \][/tex]
2. Search for the common substance:
Identify a substance that appears on both sides of the reactions. Specifically, we see that [tex]\(PCl_3\)[/tex] appears as a product in the first reaction and as a reactant in the second reaction.
3. Combine the equations:
Combine the two equations and make sure to cancel out any substances that appear on both sides:
From the first equation:
[tex]\[ 2 P (s) + 3 Cl_2 (g) \rightarrow 2 PCl_3 (l) \][/tex]
From the second equation:
[tex]\[ PCl_3 (l) + Cl_2 (g) \rightarrow PCl_5 (s) \][/tex]
In combining these, the substance [tex]\(PCl_3 (l)\)[/tex] appears on the product side in the first equation and on the reactant side in the second equation. Thus, we can cancel out [tex]\(PCl_3 (l)\)[/tex]:
[tex]\[ 2 P (s) + 3 Cl_2 (g) \rightarrow 2 PCl_3 (l) \][/tex]
[tex]\[ 2 PCl_3 (l) + 2 Cl_2 (g) \rightarrow 2 PCl_5 (s) \][/tex]
Simplifying:
[tex]\[ 2 P (s) + 3 Cl_2 (g) \rightarrow 2 PCl_3 (l) \rightarrow 2 PCl_3 (l) + 2 Cl_2 (g) \rightarrow 2 PCl_5 (s) \][/tex]
After canceling [tex]\(PCl_3 (l)\)[/tex], we get:
[tex]\[ 2 P (s) + 3 Cl_2 (g) + 2 Cl_2 (g) \rightarrow 2 PCl_5 (s) \][/tex]
Simplify further:
[tex]\[ 2 P (s) + 5 Cl_2 (g) \rightarrow 2 PCl_5 (s) \][/tex]
### Conclusion
When combining the intermediate chemical equations, the substance that gets canceled out in the overall reaction is [tex]\( PCl_3 \)[/tex].
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