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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 :

GinnyAnswer
Certainly! Let's solve this problem step by step by analyzing the given intermediate chemical equations.

### Step 1: Write Down the Equations

We are provided with 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 task is to combine these equations and determine which substance cancels out.

### Step 2: Identify Common Substances

By analyzing the two reactions, we observe that [tex]\( PCl_3 \)[/tex] is a product in the first equation and a reactant in the second. This indicates that [tex]\( PCl_3 \)[/tex] is an intermediate substance.

### Step 3: Combine the Equations

Let's write the combined equations step by step:
- First Equation: [tex]\( 2P (s) + 3Cl_2 (g) \rightarrow 2PCl_3 (l) \)[/tex]
- Second Equation: [tex]\( PCl_3 (l) + Cl_2 (g) \rightarrow PCl_5 (s) \)[/tex]

### Step 4: Balance the Intermediate Substance

We notice that [tex]\( 2PCl_3 (l) \)[/tex] is produced in the first equation and [tex]\( PCl_3 (l) \)[/tex] is consumed in the second:
- From the first equation: [tex]\( 2PCl_3 (l) \)[/tex]
- From the second equation: [tex]\( PCl_3 (l) \)[/tex]

We can balance out the mole of [tex]\( PCl_3 \)[/tex] produced and consumed.

### Step 5: Cancel Out the Intermediate Substance

If we combine the two equations while canceling out intermediate products, we get:
- [tex]\( 2P (s) + 3Cl_2 (g) \rightarrow 2PCl_3 (l) \)[/tex]
- [tex]\( 2PCl_3 (l) + 2Cl_2 (g) \rightarrow 2PCl_5 (s) \)[/tex]

Combine and cancel out one mole of [tex]\( PCl_3 \)[/tex] on each side:
- [tex]\( 2P (s) + 3Cl_2 (g) \rightarrow PCl_3 (l) + PCl_3 (l) \)[/tex]
- [tex]\( PCl_3 (l) + 2Cl_2 (g) \rightarrow PCl_5 (s) \)[/tex]

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

The substance [tex]\( PCl_3 \)[/tex] is canceled out.

### Answer

After combining the intermediate equations, the substance that is canceled out is:
[tex]\[ \boxed{PCl_3} \][/tex]
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