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]
