Question

For the reaction \[C{l_2}\left( g \right){\text{ }} + {\text{ }}KBr\left( s \right){\text{ to }}KCl\left( s \right){\text{ }} + {\text{ }}B{r_2}\left( g \right),\]how many grams of potassium chloride can be produced from $300$ grams of potassium bromide?
\[a.98.7g{\text{ }}b.111g{\text{ }}c.188g{\text{ }}d.451g\;\]

Answer 1

Hint: Potassium chloride is also known as Sylvite and also potassium salt. It is called as the metal halide salt and it is composed of the element potassium and chlorine. It is known to be odorless and has a white crystal appearance. It is known to dissolve readily in water and the solution has a salt-like taste.

Complete step-by-step answer:
Potassium chloride can be obtained from ancient dried lake deposits. It is often used in fertilizers as it has nutrient value for soils as it consists of potassium that is needed for plant growth and ion exchange. \[KCl\] is used in medicine, in scientific applications and many more areas.
For the above situation, first we need to Balance the chemical equation.
\[C{l_2}\left( g \right){\text{ }} + {\text{ }}KBr\left( s \right){\text{ }} \to {\text{ }}KCl\left( s \right){\text{ }} + {\text{ }}B{r_2}\left( g \right),\]

Then there is a need to Identify the limiting reagent. We assume it is reacting in an excess of \[C{l_2}\] (meaning \[KBr\] is our limiting reagent). Then we Convert grams to moles using \[KBr\] molar mass.
\[300{\text{ }}grams{\text{ }}KBr \cdot (1{\text{ }}mole{\text{ }}KBr/119.002{\text{ }}grams) = 2.521{\text{ }}moles{\text{ }}KBr\]

Now we need to find out how many moles of \[KCl\] can be formed be \[2.521{\text{ }}moles\]of \[KBr\] using the molar ratio between the two. This is to be done as the numbers in our balanced reaction are mole ratios, meaning $2$ moles of \[KBr\] will produce $2$moles of \[KCl\]
\[2.521{\text{ }}moles{\text{ }}KBr \cdot (2{\text{ }}moles{\text{ }}KCl/2{\text{ }}moles{\text{ }}KBr) = 2.521{\text{ }}moles{\text{ }}KCl\]

Now, Convert moles to grams using the molar mass of KCl
\[2.521{\text{ }}moles{\text{ }}KCl \cdot (74.551{\text{ }}grams/1{\text{ }}moles{\text{ }}KCl) = 187.9{\text{ }}grams{\text{ }}KCl\]

Thus, \[\left( {188{\text{ }}g} \right)\]will be the correct answer.

Note: \[KCl\] is soluble in a variety of polar solvents. In aqueous solution, it is essentially fully ionized into solvated potassium and Chlorine ions. In order to reduce it to a metal, it can be done by a reaction with metallic sodium at\[850^\circ C\]. this occurs even though potassium is more electropositive than sodium.