Question

Theoretical Yield?

K 2 CO 3 (s) + 2HCl (aq) --->2KCl (aq) + H 2 O (l) + CO 2 (g)
What would be the theoretical yield of Potassium Chloride when 41.46 g of potassium carbonate reacts with excess
hydrochloric acid?

Answer 1

The theoretical yield of `"KCl"` is 44.73 g.

Explanation:

You know that you will need a balanced chemical equation with masses, moles, and molar masses, etc.

Step 1. Assemble all the information in one place

`M_text(r):color(white)(mml)138.21color(white)(mmmmmm)74.55`
`color(white)(mmmml)"K"_2"CO"_3 + "2HCl" → "2KCl" + "H"_2"O" + "CO"_2`
`m"/g": color(white)(m.l)41.46`

Step 2. Calculate the moles of `"K"_2"CO"_3`

`"Moles of K"_2"CO"_3 = 41.46 color(red)(cancel(color(black)("mg K"_2"CO"_3))) × ("1 mol K"_2"CO"_3)/(138.21 color(red)(cancel(color(black)("g K"_2"CO"_3)))) = "0.299 97 mmol K"_2"CO"_3`

Step 3 Calculate the moles of `"KCl"`

`"Moles of KCl" = "0299 97" color(red)(cancel(color(black)("mol K"_2"CO"_3)))× ("2 mol KCl")/(1 color(red)(cancel(color(black)("mol K"_2"CO"_3)))) = "0.599 96 mmol KCl"`

Step 4. Calculate the theoretical yield of `"KCl"`

`"Mass of KCl" = "0.599 96" color(red)(cancel(color(black)("mol NaI"))) × "74.55 g KCl"/(1 color(red)(cancel(color(black)("mol KCl")))) = "44.73 g KCl"`