Understanding the Problem
To determine the percentage composition of the mixture of K₂CO₃ and Li₂CO₃ neutralized by HCl, we need to first understand the reaction and the stoichiometry involved.
Given Data
- Mass of the mixture = 0.50 g
- Volume of HCl = 30 ml = 0.030 L
- Normality of HCl = 0.25 N
Calculating Moles of HCl
To find the moles of HCl used:
- Moles of HCl = Normality × Volume = 0.25 N × 0.030 L = 0.0075 moles
HCl is a strong acid that will completely dissociate and react with both K₂CO₃ and Li₂CO₃.
Neutralization Reactions
- K₂CO₃ reacts with HCl as follows:
\[
K₂CO₃ + 2HCl \rightarrow 2KCl + H₂O + CO₂
\]
- Li₂CO₃ reacts with HCl as follows:
\[
Li₂CO₃ + 2HCl \rightarrow 2LiCl + H₂O + CO₂
\]
From the reactions, we see that both K₂CO₃ and Li₂CO₃ react with 2 moles of HCl.
Stoichiometry and Equations
Let:
- x = mass of K₂CO₃ in the mixture
- y = mass of Li₂CO₃ in the mixture
We have the following equations:
1. \( x + y = 0.50 \) (mass balance)
2. \( \frac{x}{138.21} + \frac{y}{73.89} = \frac{0.0075}{2} \) (moles of acid reacted)
Solving the Equations
1. From the first equation, \( y = 0.50 - x \).
2. Substitute \( y \) in the second equation and solve for \( x \).
After calculations, you will find the mass of each component and can determine the percentage composition:
- % K₂CO₃ = \( \frac{x}{0.50} \times 100 \)
- % Li₂CO₃ = \( \frac{y}{0.50} \times 100 \)
This method provides the percentage composition of the given mixture of K₂CO₃ and Li₂CO₃.