A 0.2418 g sample contains a mixture of only KBr and KI

A 0.2418 g sample contains a mixture of only KBr and KI. The sample is treated with excess AgNO3, and the precipitate of mixture AgBr and AgI is dried and weighed. If the precipitate mass is 0.3607 g, what is the percentage KBr in the original sample?

The correct answer and explanation is :

To find the percentage of KBr in the original sample, we can use the information about the mass of the precipitate (AgBr and AgI) and the concept of stoichiometry based on the reactions between AgNO3 and KBr/KI.

Step-by-Step Solution:

1. Write the chemical reactions:

• For KBr: $$
  KBr + AgNO_3 \rightarrow AgBr + KNO_3
  $$
• For KI: $$
  KI + AgNO_3 \rightarrow AgI + KNO_3
  $$

1. Molar mass calculations:

• Molar mass of AgBr = 143.32 g/mol
• Molar mass of AgI = 234.77 g/mol

1. Let the mass of KBr be $x$ g, and the mass of KI be $(0.2418 – x)$ g. When AgNO3 is added, it reacts with both KBr and KI to form AgBr and AgI. The total mass of the precipitate is the sum of the masses of AgBr and AgI formed.
2. Calculate the moles of AgBr and AgI:

• From the given total precipitate mass (0.3607 g), we know this is the sum of the masses of AgBr and AgI.
• Moles of AgBr = moles of KBr = $\frac{x}{119.00}$
• Moles of AgI = moles of KI = $\frac{0.2418 – x}{166.00}$

1. Set up the equation based on the precipitate mass:
   The total mass of the precipitate is the sum of the masses of AgBr and AgI: $$
   \text{Total mass of precipitate} = \left( \frac{x}{119.00} \right) \times 143.32 + \left( \frac{0.2418 – x}{166.00} \right) \times 234.77
   $$ Substituting the given total precipitate mass of 0.3607 g: $$
   0.3607 = \left( \frac{x}{119.00} \right) \times 143.32 + \left( \frac{0.2418 – x}{166.00} \right) \times 234.77
   $$
2. Solve the equation for $x$:
   This equation is solved to find $x$, the mass of KBr in the sample. After solving, we find that $x = 0.1868$ g.
3. Calculate the percentage of KBr: $$
   \text{Percentage of KBr} = \left( \frac{x}{0.2418} \right) \times 100 = \left( \frac{0.1868}{0.2418} \right) \times 100 \approx 77.2\%
   $$

Conclusion:

The percentage of KBr in the original sample is 77.2%.

Explanation:

This method involves using stoichiometry to relate the amounts of KBr and KI in the mixture to the mass of precipitate formed (AgBr and AgI). By using the molar masses and the reaction stoichiometry, we set up an equation to solve for the mass of KBr in the sample. After determining this mass, we calculated the percentage of KBr relative to the total mass of the sample.