Answer:
The correct answer is ;
Cancel out CO because it appears as a reactant in one intermediate reaction and a product in the other intermediate reaction.
The final chemical reaction:
[tex]C(s)+1O_2+\rightarrow CO_2(g)[/tex]
Explanation:
Step 1 :
[tex]C(s)+\frac{1}{2}O_2(g)\rightarrow CO(g)[/tex]..[1]
Step 2:
[tex]CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)[/tex]..[2]
Adding equation [1] and [2] will give us final equation :
[tex]C(s)+\frac{1}{2}O_2(g)+\frac{1}{2}O_2(g)+CO(g)\rightarrow CO(g)+CO_2(g)[/tex]
CO (g) on both sides will get cancel and the final equation comes out to be :
[tex]C(s)+1O_2+\rightarrow CO_2(g)[/tex]
CO as a reactant and a product in the final chemical reaction is written as:
[tex]C(s)+1O_2(g)+CO(g)\rightarrow CO_2(g)+(CO(g)[/tex]
Since CO has been written in both the reactant and the product side, CO has been the intermediate. In the final reaction, the intermediate has been canceled out.
The given intermediate chemical reactions are:
First reaction : [tex]\rm C\;_(_s_)\;+\;\dfrac{1}{2}\;O_2\;\rightarrow\;CO\;_(_g_)[/tex]
Second reaction: [tex]\rm CO\;+\;\dfrac{1}{2}\; O_2\;_(_g_)\;\rightarrow\;CO_2\;_(_g_)[/tex]
From the given intermediate reactions, the final reaction will be:
[tex]\rm C\;_(_g_)\;+\;O_2\;+\;CO\;_(_g_)\;\rightarrow\;CO\;_(_g_)\;+\;CO_2[/tex]
Since CO has been written in both the reactant and the product side, CO has been the intermediate. In the final reaction, the intermediate has been canceled out. Thus the final reaction will be as follows:
[tex]\rm C\;_(_g_)\;+\;O_2\;\rightarrow\;CO_2[/tex]
The reaction with CO as a reactant and a product in the final chemical reaction will be:
[tex]\rm C\;_(_g_)\;+\;O_2\;+\;CO\;_(_g_)\;\rightarrow\;CO\;_(_g_)\;+\;CO_2[/tex].
For more information about the intermediate equations, refer to the link:
https://brainly.com/question/12790981
