Question

Given the following equation: 2K + Cl2 -> 2KCl How many grams of KCl is produced from 4.00 g of K and excess Cl2?

Answer 1

Answer:

42.65g

Explanation:

Given parameters:

Mass of K = 4g

Unknown: Mass of KCl

Solution:

  Complete equation of the reaction:

              2K + Cl₂ → 2KCl

To solve this problem, we know that the reactant in short supply is potassium K and this dictates the amount of products that would be formed. The chlorine gas is in excess and we can't use it to determine the amount of product that would form.

Now, we work from the known to the unknown. Since we know the mass of K given in the reaction, we can simply find the molar relationship between the reacting potassium and the product. We simply convert the mass to mole and compare to the product. From there we can find the mass of KCl that would be produced.

Calculating number of moles of K

      Number of moles = $\frac{mass}{molar mass}$

        Number of moles of K =  $\frac{4}{39}$ = 0.103mol

From the given reaction equation:

   2 moles of K will produce 2 moles of KCl

 Therefore 0.103mol of K will produce 0.103mol of KCl

To find the mass of KCl produced,

   Mass of KCl = number of moles of KCl x molar mass

Molar mass of KCl = 39 + 35.5 = 74.5gmol⁻¹

Mass of KCl = 0.103 x 74.5 = 42.65g

Answer 2

Answer:

Grams of KCl produced = 7.60 g

Explanation:

The given reaction is:

2K + Cl2 → 2KCl

It is given that Cl2 is in excess which implies that potassium K is the limiting reagent

Based on the reaction stoichiometry:

2 moles of K produces 2 moles of KCl i.e the molar ratio of K:KCl = 1:1

$Moles(K)= \frac{Mass}{Atomic mass}=\frac{4.00g}{39.09g/mol}=0.102moles$

Therefore, moles of KCl produced = 0.102

$Mass(KCl)=moles*mol.wt = 0.102moles*74.55g/mol=7.60g$