Question

A 50-g sample of impure \(\mathrm{KClO}_{3}\) (solubility \(=7.1 \mathrm{~g}\) per \(100 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\) at \(\left.20^{\circ} \mathrm{C}\right)\) is contaminated with 10 percent of \(\mathrm{KCl}\) (solubility \(=25.5 \mathrm{~g}\) per \(100 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) at \(\left.20^{\circ} \mathrm{C}\right)\) Calculate the minimum quantity of \(20^{\circ} \mathrm{C}\) water needed to dissolve all the \(\mathrm{KCl}\) from the sample. How much \(\mathrm{KClO}_{3}\) will be left after this treatment? (Assume that the solubilities are unaffected by the presence of the other compound.)

Short answer

19.61 g of water is needed; 43.61 g of KClO3 remains.

Step-by-step solution

Determine the Mass of KCl in the Sample

The sample contains 10% KCl, so the mass of KCl in the 50-g sample is calculated by multiplying 0.10 by 50 g. This gives:\[ \text{Mass of KCl} = 0.10 \times 50 \, \text{g} = 5 \, \text{g} \]

Calculate the Minimum Water Required to Dissolve KCl

Given the solubility of KCl is 25.5 g per 100 g of water, calculate the amount of water needed to dissolve 5 g of KCl. Use a proportion:\[ \frac{5 \, \text{g}}{25.5 \, \text{g}} = \frac{x \, \text{g of water}}{100 \, \text{g of water}} \]Solving for \(x\), \[ x = \frac{5 \, \text{g} \times 100 \, \text{g}}{25.5 \, \text{g}} \approx 19.61 \, \text{g of water} \]

Determine the Mass of KClO3 in the Sample

The remaining 90% of the sample is KClO3. Calculate the mass of KClO3:\[ \text{Mass of KClO}_{3} = 0.90 \times 50 \, \text{g} = 45 \, \text{g} \]

Calculate How Much KClO3 is Left After Treatment

Since the solubility of KClO3 at 20°C is 7.1 g per 100 g of water, calculate how much KClO3 would dissolve in the calculated 19.61 g of water:\[ \text{Using proportion} \]\[ \frac{7.1 \, \text{g KClO}_3}{100 \, \text{g water}} = \frac{x \, \text{g KClO}_3}{19.61 \, \text{g water}} \]Solving for \(x\):\[ x = \frac{7.1 \, \text{g} \times 19.61 \, \text{g}}{100 \, \text{g}} \approx 1.39 \, \text{g of KClO}_3 \]Therefore, the mass of KClO3 left after treatment is:\[ 45 \, \text{g} - 1.39 \, \text{g} \approx 43.61 \, \text{g} \]

Key concepts

KClO3

Potassium chlorate (\(\text{KClO}_3\)) is a chemical compound known for its use in oxygen production and as a strong oxidizer in fireworks and explosives. It decomposes readily to produce oxygen when heated and is a key component in safety matches. In this problem, we are concerned with its solubility, which refers to how much of the substance can dissolve in a certain amount of solvent (water) at a specific temperature. \(\text{KClO}_3\)'s solubility at \(20^\circ \text{C}\) is relatively low at 7.1 g per 100 g of water. This solubility is crucial to calculations when determining how much of it will remain in a mixture after dissolution treatment. In context, if we mix \(\text{KClO}_3\)with water, only a limited amount will dissolve at this temperature, and excess will remain as solid, influencing the percentage left after treatment.

KCl

Potassium chloride (\(\text{KCl}\)) is a well-known mineral with significant roles in fertilizer production and as a salt substitute in food. In this context, it's an impurity in the \(\text{KClO}_3\)sample. The solubility of \(\text{KCl}\)is higher compared to \(\text{KClO}_3\), with 25.5 g able to dissolve in 100 g of water at \(20^\circ \text{C}\).
This factor becomes critical when determining how much water is needed to dissolve it completely from the mixture. First, calculating the mass of \(\text{KCl}\)in the impure sample gives 5 g, considering it makes up 10% of the 50-g sample.
Then, using its solubility ratio, one can compute the exact volume of water required to dissolve all the \(\text{KCl}\)and ensure pure \(\text{KClO}_3\)remains.

impure sample

An impure sample contains more than one substance, with impurities possibly affecting reactions, solubility, and other physical or chemical properties. Here, we have a mixed sample of \(\text{KClO}_3\)and \(\text{KCl}\), with the latter being the impurity.
This composition significantly impacts how much of each component dissolves when treated with water. In analytical processes or industrial applications, one often encounters such mixtures, requiring careful calculations to restore purity.
For instance, separating \(\text{KClO}_3\)from \(\text{KCl}\)relies on understanding and leveraging their different solubility properties. Through this approach, the impure sample undergoes separation, enabling accurate assessment of final masses, like the 43.61 g of \(\text{KClO}_3\)left after the dissolution process.

mass calculation

Mass calculation is a core aspect of chemistry, helping determine precise amounts in reactions and solutions. For our problem, it involves calculating parts of a mixture and the solvents required for dissolution. The process starts with calculating percentages based on known impurities, like the 10% \(\text{KCl}\) in the sample, leading to identifying 45 g of \(\text{KClO}_3\) present.
Next, understanding solubility translates percentages into potential amounts dissolved in a specific water volume. Based on the solubility ratio, the water needed to dissolve the 5 g of \(\text{KCl}\) is approximately 19.61 g. These precise calculations ensure each component, like \(\text{KClO}_3\), is duly accounted for after treatment, leading to 43.61 g remaining, showcasing the combined importance of thorough mass and solubility understanding.