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Consider the following intermediate chemical equations:

[tex]\[
\begin{array}{l}
2P(s) + 3Cl_2(g) \rightarrow 2PCl_3(l) \\
PCl_3(l) + Cl_2(g) \rightarrow PCl_5(s)
\end{array}
\][/tex]

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

A. [tex]\( P \)[/tex]
B. [tex]\( Cl_2 \)[/tex]
C. [tex]\( PCl_3 \)[/tex]
D. [tex]\( PCl_5 \)[/tex]

Sagot :

GinnyAnsweridn
To determine which substance cancels out when combining the chemical equations, let's go through a detailed step-by-step process.

We are given the following intermediate chemical equations:

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

Our goal is to combine these two equations and identify the substance that cancels out.

Step 1: Write down the reactions

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

Step 2: Combine the equations

To do this, we need to add the two reactions together. We must ensure that we have the same number of each type of atom on both sides of the combined equation.

First, list all reactants and products before combining:

- Reactants: [tex]\(2P(s) + 3Cl_2(g) + PCl_3(l) + Cl_2(g)\)[/tex]
- Products: [tex]\(2PCl_3(l) + PCl_5(s)\)[/tex]

Since [tex]\(PCl_3(l)\)[/tex] appears on both the reactant and product sides, we need to adjust the alignment for combination. We can rewrite as:

[tex]\[ 2P(s) + 3Cl_2(g) \rightarrow 2PCl_3(l) \quad (Equation \,1) \][/tex]
[tex]\[ PCl_3(l) + Cl_2(g) \rightarrow PCl_5(s) \quad (Equation \,2) \][/tex]

Step 3: Adjust and Combine Reactants and Products

Aligning and summing both equations, we get:

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

Combining the equations, keep track of what cancels out. The term [tex]\(PCl_3(l)\)[/tex] appears on both sides:

On the left: [tex]\(2P(s) + 4Cl_2(g) \rightarrow 2PCl_3(l) + Cl_2(g) \rightarrow PCl_5(s)\)[/tex]

Simplifying further, the [tex]\(PCl_3(l)\)[/tex] on the reactant side and [tex]\(PCl_3(l)\)[/tex] on the product side cancel each other out:

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

Step 4: Identify the cancelling substance

After canceling the common [tex]\(PCl_3(l)\)[/tex], we are left with:

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

So, the substance that cancels out in the combined equations is [tex]\(PCl_3(l)\)[/tex].

Thus, the substance that gets canceled out is:

[tex]\(PCl_3\)[/tex]
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