Answer :
To determine the substance that cancels out when combining the intermediate chemical equations, let's carefully examine the given reactions.
We are given two 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]
The goal is to combine these equations and identify the common substance that appears in both the products of the first reaction and the reactants of the second reaction, as this will be the substance that cancels out.
1. The first equation suggests:
[tex]\[2P(s) + 3Cl_2(g) \rightarrow 2PCl_3(l)\][/tex]
2. The second equation suggests:
[tex]\[PCl_3(l) + Cl_2(g) \rightarrow PCl_5(s)\][/tex]
Upon combining these two equations, let's align them step-by-step:
- In the first equation, [tex]\(2PCl_3(l)\)[/tex] is produced.
- In the second equation, [tex]\(PCl_3(l)\)[/tex] is consumed.
When writing the overall reaction, we need to ensure that the number of molecules of each substance produced in one reaction equates to the same number of molecules consumed in the next reaction.
By adding the two equations, the intermediate product [tex]\(PCl_3(l)\)[/tex] cancels out because it is produced and then used up in the subsequent reaction.
1. [tex]\(2P(s) + 3Cl_2(g) \rightarrow 2PCl_3(l)\)[/tex]
2. [tex]\(2PCl_3(l) + 2Cl_2(g) \rightarrow 2PCl_5(s)\)[/tex]
Combine:
[tex]\[2P(s) + 5Cl_2(g) \rightarrow 2PCl_5(s)\][/tex]
Therefore, the intermediate substance [tex]\(PCl_3(l)\)[/tex] is the one that cancels out when the intermediate chemical equations are combined.
Thus, the substance that cancels out is:
[tex]\[ \boxed{PCl_3} \][/tex]
We are given two 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]
The goal is to combine these equations and identify the common substance that appears in both the products of the first reaction and the reactants of the second reaction, as this will be the substance that cancels out.
1. The first equation suggests:
[tex]\[2P(s) + 3Cl_2(g) \rightarrow 2PCl_3(l)\][/tex]
2. The second equation suggests:
[tex]\[PCl_3(l) + Cl_2(g) \rightarrow PCl_5(s)\][/tex]
Upon combining these two equations, let's align them step-by-step:
- In the first equation, [tex]\(2PCl_3(l)\)[/tex] is produced.
- In the second equation, [tex]\(PCl_3(l)\)[/tex] is consumed.
When writing the overall reaction, we need to ensure that the number of molecules of each substance produced in one reaction equates to the same number of molecules consumed in the next reaction.
By adding the two equations, the intermediate product [tex]\(PCl_3(l)\)[/tex] cancels out because it is produced and then used up in the subsequent reaction.
1. [tex]\(2P(s) + 3Cl_2(g) \rightarrow 2PCl_3(l)\)[/tex]
2. [tex]\(2PCl_3(l) + 2Cl_2(g) \rightarrow 2PCl_5(s)\)[/tex]
Combine:
[tex]\[2P(s) + 5Cl_2(g) \rightarrow 2PCl_5(s)\][/tex]
Therefore, the intermediate substance [tex]\(PCl_3(l)\)[/tex] is the one that cancels out when the intermediate chemical equations are combined.
Thus, the substance that cancels out is:
[tex]\[ \boxed{PCl_3} \][/tex]
