Question

3. If you start with 8x1025 molecules of Cl, and 25 grams of KI, how many grams of KCl would

be made? Calculate the percent yield if 10.0g of KCl is actually produced.

Answer 1

Percent yield = 89.1%

Step-by-step explanation:

Based on the equation:

Cl₂ + 2KI → 2KCl + I₂

1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl

To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:

Moles Cl₂:

8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles

Moles KI -Molar mass: 166.0028g/mol-

25g * (1mol / 166.0028g) = 0.15 moles

Here, clarely, the KI is the limiting reactant

As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:

0.15 moles * (74.5513g / mol) =

11.2g KCl

Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100

Percent yield = 89.1%