# 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?

Jun 18, 2018

The theoretical yield of $\text{KCl}$ 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

${M}_{\textrm{r}} : \textcolor{w h i t e}{m m l} 138.21 \textcolor{w h i t e}{m m m m m m} 74.55$
$\textcolor{w h i t e}{m m m m l} {\text{K"_2"CO"_3 + "2HCl" → "2KCl" + "H"_2"O" + "CO}}_{2}$
$m \text{/g} : \textcolor{w h i t e}{m . l} 41.46$

Step 2. Calculate the moles of ${\text{K"_2"CO}}_{3}$

${\text{Moles of K"_2"CO"_3 = 41.46 color(red)(cancel(color(black)("mg K"_2"CO"_3))) × ("1 mol K"_2"CO"_3)/(138.21 color(red)(cancel(color(black)("g K"_2"CO"_3)))) = "0.299 97 mmol K"_2"CO}}_{3}$

Step 3 Calculate the moles of $\text{KCl}$

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

Step 4. Calculate the theoretical yield of $\text{KCl}$

$\text{Mass of KCl" = "0.599 96" color(red)(cancel(color(black)("mol NaI"))) × "74.55 g KCl"/(1 color(red)(cancel(color(black)("mol KCl")))) = "44.73 g KCl}$