Simple Chemistry HW Problem: Calculating Composition of a KCl and KBr Mixture

[ChickenChakuro]

Simple? Chemistry Problem

Hello everyone, have a quick chemistry problem I'm having trouble figuring out:

Q. A sample is a mixture of KCl and KBr. When 0.1024 g of the sample is dissolved in water and reacted with excess silver nitrate, 0.1889 g solid is obtained. What is the composition by mass percent of the original mixture?

I'm braindead. I can't even come up with the original equation...I was thinking at first that I have to split it into two equations,

$$KCl + AgNO_3 \rightarrow AgCl + KNO_3$$

$$KBr + AgNO_3 \rightarrow AgBr + KNO_3$$

But I'm not sure what to do next. Perhaps do mass composition of Chloride and Bromide ions? Ack. Please help!

 

[lippyka]

Answer: The first step is to calculate the amount of KCl and KBr in the sample. We can do this using the given amounts of the sample and the product: Mass of KCl = 0.1889g - 0.1024g = 0.0865g Mass of KBr = 0.1024g - 0.0865g = 0.0159g Now we can calculate the composition by mass percent of the original mixture: KCl mass percent = (0.0865g/0.1024g) x 100% = 84.5% KBr mass percent = (0.0159g/0.1024g) x 100% = 15.5% Therefore, the composition by mass percent of the original mixture is 84.5% KCl and 15.5% KBr.

 

[quicknote]

Hi there, this is a great question and I can see why you might find it challenging. Let's break it down step by step.

First, we need to understand what is happening in the reaction. When a mixture of KCl and KBr is reacted with silver nitrate, the silver ions (Ag+) will react with the chloride (Cl-) and bromide (Br-) ions to form insoluble silver chloride (AgCl) and silver bromide (AgBr) precipitates. This means that the remaining ions, potassium (K+) and nitrate (NO3-), will remain in solution.

Next, we can write out the balanced chemical equation for the reaction:

KCl + AgNO3 -> AgCl + KNO3

KBr + AgNO3 -> AgBr + KNO3

Now, let's look at the given information. We know that 0.1024 g of the mixture was used and 0.1889 g of solid was obtained. We also know that the solid is a mixture of AgCl and AgBr in a 1:1 ratio, meaning that the mass of AgCl and AgBr are equal.

Using this information, we can set up a system of equations to solve for the mass of each component in the original mixture:

0.1024 g mixture = x g KCl + y g KBr (where x and y are the masses of KCl and KBr, respectively)

0.1889 g solid = x g AgCl + y g AgBr (since the mass of AgCl and AgBr are equal)

Solving this system of equations, we get x = 0.075 g and y = 0.0279 g. This means that the original mixture contained 0.075 g of KCl and 0.0279 g of KBr.

To find the mass percent composition of each component, we divide the mass of each component by the total mass of the mixture and multiply by 100%:

Mass percent KCl = (0.075 g / 0.1024 g) * 100% = 73.24%

Mass percent KBr = (0.0279 g / 0.1024 g) * 100% = 27.19%

Therefore, the composition by mass percent of the original mixture is 73.24% KCl and 27.19% KBr.

I hope this helps clarify the problem