Question

Chlorine may be prepared by the action of \(\mathrm{KClO}_{3}\) on \(\mathrm{HCl}\), and the reaction may be represented by the equation: \(\mathrm{KClO}_{3}+6 \mathrm{HCl} \rightarrow \mathrm{KCl}+3 \mathrm{Cl}_{2}+3 \mathrm{H}_{2} \mathrm{O}\) Calculate the weight of \(\mathrm{KClO}_{3}\) which would be required to produce \(1.0\) liter of \(\mathrm{Cl}_{2}\) gas at STP. \(\mathrm{R}=.082\) liter-atm/mole \(-{ }^{\circ} \mathrm{K}\)

Short answer

To produce 1.0 L of Cl₂ gas at STP, you would need approximately 7.64 g of KClO₃.

Step-by-step solution

Determine the Number of Moles of Cl2 from the Volume

Utilize the Ideal Gas Law in the form \(n=\frac{PV}{RT}\) to calculate the number of moles of \(Cl_{2}\): Here, P (pressure) = 1 atm (standard temperature and pressure), V (volume) = 1 L, R = 0.082 L atm/mol K, and T (temperature) = 273.15 K (which is the temperature at standard conditions, 0 degree Celsius). Substitute all given values into the formula to find the number of moles of Cl2.

Use Stoichiometry to Determine the Required Moles of KClO3

According to the stoichiometry of the reaction, one mole of \(\mathrm{KClO}_{3}\) produces three moles of \(\mathrm{Cl}_{2}\). Therefore, determine the number of moles of \(\mathrm{KClO}_{3}\) needed to produce the calculated number of moles of \(\mathrm{Cl}_{2}\) in Step 1, using the mole ratio from the balanced chemical equation.

Convert the Number of Moles of KClO3 to Weight

Use the molar mass of \(\mathrm{KClO}_{3}\) (which is approximately 122.55 g/mol) to convert the number of moles of \(\mathrm{KClO}_{3}\) calculated in Step 2 into weight. The needed amount of \(\mathrm{KClO}_{3}\) is the product of the calculated number of moles and the molar mass of \(\mathrm{KClO}_{3}\). The molar mass is calculated by adding the molar mass of Potassium, Chlorine and three times Oxygen. This calculation yields the weight of \(\mathrm{KClO}_{3}\) which would be required to produce 1.0 liter of \(\mathrm{Cl}_{2}\) gas at STP. Remember to use g/mol as the unit for the calculated weight.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry used to relate the pressure, volume, temperature, and number of moles of a gas. It is expressed as \( PV = nRT \), where:

• \( P \) represents pressure of the gas in atmospheres (atm),
• \( V \) stands for the volume of the gas in liters (L),
• \( n \) is the number of moles of the gas,
• \( R \) is the ideal gas constant \(0.082\) L atm/mol K,
• \( T \) is the temperature in Kelvin (K).

This formula allows you to calculate any one of the parameters if the others are known. By rearranging the formula to \( n = \frac{PV}{RT} \), you can solve for the number of moles \( n \) when the gas is at Standard Temperature and Pressure (STP), where \( P = 1 \) atm and \( T = 273.15 \) K.

Chemical Reaction

Chemical reactions involve the transformation of reactants into products. In the provided exercise, the reaction involves \( \mathrm{KClO}_{3} \) and \( \mathrm{HCl} \) to produce \( \mathrm{KCl} \), \( \mathrm{Cl}_{2} \), and \( \mathrm{H}_{2} \mathrm{O} \). This is a typical example of a redox reaction where:

• \( \mathrm{KClO}_{3} \) acts as an oxidant and \( \mathrm{HCl} \) serves as a reducing agent.
• The balanced equation demonstrates the stoichiometry by showing the exact number of moles of each reactant and product.

The stoichiometric coefficients indicate the mole ratios. In this equation, one mole of \( \mathrm{KClO}_{3} \) is needed for three moles of \( \mathrm{Cl}_{2} \), which means it produces three times the amount of chlorine gas compared to its own amount by moles.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). Understanding molar mass is crucial for converting moles to grams and vice versa. For the reactant \( \mathrm{KClO}_{3} \) in this exercise, the molar mass is approximately 122.55 g/mol.

• Molar mass can be calculated by summing the atomic masses of all atoms in the formula.
• For \( \mathrm{KClO}_{3} \), it consists of Potassium (\( \approx 39.10 \) g/mol), Chlorine (\( \approx 35.45 \) g/mol), and three Oxygens (\(3 \times \approx 16.00\) g/mol).

This knowledge allows you to convert between the number of moles and the weight of \( \mathrm{KClO}_{3} \), enabling precise reactions and calculations.

Standard Temperature and Pressure

Standard Temperature and Pressure (STP) is a reference point used in chemistry to ensure consistency in experiments and calculations. At STP, the conditions are:

• Pressure at \(1\) atm (atmosphere),
• Temperature at \(273.15\) Kelvin (K), equivalent to \(0\) degrees Celsius.

Using STP simplifies the computation involving gases, as these are standard conditions at which the properties of gases have been widely tabulated. Essentially, STP provides a standard "baseline" for gas calculations, avoiding discrepancies due to varying environmental conditions. This is why it is frequently used in conjunction with the Ideal Gas Law to determine the amount of gas products or reactants in chemical reactions.