Answer:
* Limiting reagent: potassium.
* [tex]m_{Cl_2}^{excess}=2.47gCl_2[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]2K(s)+Cl_2(g)\rightarrow 2KCl[/tex]
In such a way, the reacting moles of chlorine are:
[tex]n_{Cl_2}=7.00g*\frac{1molCl_2}{70.9gCl_2}=0.0987molCl_2[/tex]
Now, the amount of chlorine gas that would react with 5.00 g of potassium result:
[tex]n_{Cl_2}^{consumed}=5.00g*\frac{1molK}{39.1gK}*\frac{1molCl_2}{2molK}=0.0639molCl_2[/tex]
Thus, the chlorine will be in excess and the potassium will be the limiting reagent. Therefore, the mass of excess chlorine turns out:
[tex]m_{Cl_2}^{excess}=(0.0987-0.0639)molCl_2*\frac{70.9gCl_2}{1molCl_2}=2.47gCl_2[/tex]
Best regards.
