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Chapter 3: Problem 103

A mixture of potassium chloride and potassium bromide weighing \(3.595 \mathrm{~g}\) is heated with chlorine, which converts the mixture completely to potassium chloride. The total mass of potassium chloride after the reaction is \(3.129 \mathrm{~g}\). What percentage of the original mixture was potassium bromide?

Short Answer

Expert verified
Answer: The percentage of potassium bromide (KBr) in the original mixture is approximately 19.3%.

Step by step solution

01

Calculate the mass difference before and after chlorination

As the mass of the mixture before the reaction is given as \(3.595 \mathrm{~g}\), and after the reaction, it is \(3.129 \mathrm{~g}\), we can find the mass difference by subtracting the mass after from the mass before: $$ \text{Mass difference} = 3.595 - 3.129 = 0.466 \mathrm{~g} $$ Now, we know that the mass difference (\(0.466 \mathrm{~g}\)) is the mass of bromine (Br) replaced by chlorine (Cl) during the reaction.
02

Calculate the number of moles of bromine (Br) replaced

We can now calculate the number of moles of bromine (Br) that were replaced by chlorine (Cl). We will use the molar mass of bromine (Br), which is approximately \(79.90 \mathrm{~g/mol}\): $$ \text{Moles of Br} = \frac{\text{Mass of Br}}{\text{Molar mass of Br}} = \frac{0.466}{79.90} \approx 0.00583 \mathrm{~mol} $$ So, \(0.00583\) moles of bromine (Br) were replaced during the reaction.
03

Calculate the mass of potassium bromide (KBr) in the original mixture

Now, we can calculate the mass of potassium bromide (KBr) in the original mixture. We will use the molar mass of potassium bromide (KBr), which is approximately \(39.10 \mathrm{~g/mol} + 79.90 \mathrm{~g/mol} = 119.00 \mathrm{~g/mol}\): $$ \text{Mass of KBr in the original mixture} = \text{Moles of Br} \times \text{Molar mass of KBr} \approx 0.00583 \times 119.00 \approx 0.693 \mathrm{~g} $$ So, there are \(0.693 \mathrm{~g}\) of potassium bromide (KBr) in the original mixture.
04

Calculate the percentage of potassium bromide (KBr) in the original mixture

Finally, we can calculate the percentage of potassium bromide (KBr) in the original mixture, using the mass of potassium bromide (KBr) we found in Step 3 and the total initial mass of the mixture: $$ \text{Percentage of KBr} = \frac{\text{Mass of KBr}}{\text{Initial mass of the mixture}} \times 100 \%= \frac{0.693}{3.595} \times 100 \% \approx 19.3 \% $$ So, approximately \(19.3 \%\) of the original mixture was potassium bromide (KBr).

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