Question

Potassium chloride is used as a substitute for sodium chloride for individuals with high blood pressure. Identify the limiting reactant and determine the mass of the excess reactant remaining when 7.00 g of chlorine gas reacts with 5.00 g of potassium to form potassium chloride.

Answer 1

Answer:

* Limiting reagent: potassium.

* $m_{Cl_2}^{excess}=2.47gCl_2$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2K(s)+Cl_2(g)\rightarrow 2KCl$

In such a way, the reacting moles of chlorine are:

$n_{Cl_2}=7.00g*\frac{1molCl_2}{70.9gCl_2}=0.0987molCl_2$

Now, the amount of chlorine gas that would react with 5.00 g of potassium result:

$n_{Cl_2}^{consumed}=5.00g*\frac{1molK}{39.1gK}*\frac{1molCl_2}{2molK}=0.0639molCl_2$

Thus, the chlorine will be in excess and the potassium will be the limiting reagent. Therefore, the mass of excess chlorine turns out:

$m_{Cl_2}^{excess}=(0.0987-0.0639)molCl_2*\frac{70.9gCl_2}{1molCl_2}=2.47gCl_2$

Best regards.