Question

A mixture contained zinc sulfide, \(\mathrm{ZnS},\) and lead sulfide, PbS. A sample of the mixture weighing \(6.12 \mathrm{~g}\) was reacted with an excess of hydrochloric acid. The reactions are $$ \mathrm{ZnS}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{~S}(g) $$ \(\mathrm{PbS}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{PbCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{~S}(g)\) If the sample reacted completely and produced \(1.049 \mathrm{~L}\) of hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\), at \(23^{\circ} \mathrm{C}\) and \(762 \mathrm{mmHg},\) what were the percentages of \(\mathrm{ZnS}\) and \(\mathrm{PbS}\) in the mixture?

Short answer

47.34% ZnS and 52.66% PbS in the mixture.

Step-by-step solution

- Convert Volume to Moles

To find the moles of \( \mathrm{H}_2\mathrm{S} \), use the ideal gas law: \( PV = nRT \). First, convert the temperature from Celsius to Kelvin: \( 23^\circ C = 23 + 273.15 = 296.15 \ K \). Convert pressure from mmHg to atm: \( 762 \ \mathrm{mmHg} = \frac{762}{760} \ \mathrm{atm} \approx 1.0026 \ \mathrm{atm} \). The volume given is \( 1.049 \ \mathrm{L} \). Use the gas constant \( R = 0.0821 \ \mathrm{L \ atm \ mol^{-1} \ K^{-1}} \). Substitute these values into the equation: \( n = \frac{PV}{RT} = \frac{1.0026 \times 1.049}{0.0821 \times 296.15} \approx 0.0432 \ \mathrm{mol} \).

- Set Up System of Equations

Since \( \mathrm{ZnS} \) and \( \mathrm{PbS} \) both produce \( \mathrm{H}_2\mathrm{S} \) in a 1:1 molar ratio with their reaction, the moles of \( \mathrm{H}_2\mathrm{S} \) from each compound equal the moles of the compound reacted. Let \( x \) be the moles of \( \mathrm{ZnS} \) and \( y \) be the moles of \( \mathrm{PbS} \). Then: \( x + y = 0.0432 \) (from total moles of \( \mathrm{H}_2\mathrm{S} \) produced).

- Mass Balance Equation

The sample total mass is given as \( 6.12 \ \mathrm{g} \). Convert moles to mass for \( \mathrm{ZnS} \) (molar mass \( 97.47 \ \mathrm{g/mol} \)) and \( \mathrm{PbS} \) (molar mass \( 239.26 \ \mathrm{g/mol} \)), forming the equation: \( 97.47x + 239.26y = 6.12 \).

- Solve the System of Equations

We have two equations: \( x + y = 0.0432 \) and \( 97.47x + 239.26y = 6.12 \). Solve for \( x \) in terms of \( y \): \( x = 0.0432 - y \). Substitute \( x \) in the mass balance equation: \( 97.47(0.0432 - y) + 239.26y = 6.12 \). Simplifying gives: \( 4.2118 - 97.47y + 239.26y = 6.12 \), leading to \( 141.79y = 1.9082 \), thus \( y = \frac{1.9082}{141.79} \approx 0.0135 \ \mathrm{mol} \). Substituting \( y \) back to find \( x \): \( x = 0.0432 - 0.0135 = 0.0297 \ \mathrm{mol} \).

- Calculate Percentages

Convert moles back to mass to find composition. Mass of \( \mathrm{ZnS} \) is \( 0.0297 \times 97.47 = 2.895 \ \mathrm{g} \), and mass of \( \mathrm{PbS} \) is \( 0.0135 \times 239.26 = 3.224 \ \mathrm{g} \). Calculate the percentage of each in the mixture: \( \% \ \mathrm{ZnS} = \frac{2.895}{6.12} \times 100 = 47.34\% \), \( \% \ \mathrm{PbS} = \frac{3.224}{6.12} \times 100 = 52.66\% \).

Key concepts

Ideal Gas Law

The Ideal Gas Law is an essential equation in chemistry that helps us understand the relationship between pressure, volume, temperature, and the number of moles of a gas. The formula is expressed as \( PV = nRT \), where \( P \) stands for pressure, \( V \) represents volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) represents temperature in Kelvin.

To solve a problem like the one presented in the exercise, you begin by converting all given conditions like temperature and pressure into the correct units. In our example, temperature in Celsius is converted to Kelvin by adding 273.15, and pressure in mmHg is converted to atmospheres by dividing by 760. These adjustments ensure that your values are compatible with the constant \( R = 0.0821 \, \text{L atm/mol K} \).

By plugging all these values into the Ideal Gas Law formula, you can solve for \( n \), the number of moles of gas, which in this case is hydrogen sulfide \( \mathrm{H}_2\mathrm{S} \). This foundational equation is crucial when dealing with stoichiometry in gas reactions.

Molar Mass

Molar mass is a key concept to master when dealing with chemical reactions and stoichiometry. It is defined as the mass of one mole of a substance and is expressed in \( \text{g/mol} \). For the compounds in our exercise, the molar mass of zinc sulfide \( \mathrm{ZnS} \) and lead sulfide \( \mathrm{PbS} \) are 97.47 \( \text{g/mol} \) and 239.26 \( \text{g/mol} \) respectively.

The exercise involves converting moles of \( \mathrm{ZnS} \) and \( \mathrm{PbS} \) into mass to solve a mixture problem. This is done using the equation \( \text{mass} = \text{moles} \times \text{molar mass} \). By finding the mass of each compound in the mixture, you can determine the proportion by mass of each component in the total sample.

Understanding molar mass allows you to convert between moles and grams, bridging the gap between the macroscopic world of reactions and the microscopic world of molecules and atoms.

Chemical Reactions

Chemical reactions describe the process where substances, known as reactants, are transformed into different substances, known as products. In our specific exercise, zinc sulfide \( \mathrm{ZnS} \) and lead sulfide \( \mathrm{PbS} \) both react with hydrochloric acid \( \mathrm{HCl} \) to produce hydrogen sulfide \( \mathrm{H}_2 \mathrm{S} \) gas. The reactions follow these equations:

• \( \mathrm{ZnS}(s) + 2\mathrm{HCl}(aq) \longrightarrow \mathrm{ZnCl}_2(aq) + \mathrm{H}_2 \mathrm{S}(g) \)
• \( \mathrm{PbS}(s) + 2\mathrm{HCl}(aq) \longrightarrow \mathrm{PbCl}_2(aq) + \mathrm{H}_2 \mathrm{S}(g) \)

Both reactions demonstrate a 1:1 molar relationship between the metal sulfides and hydrogen sulfide, which is significant for calculating the amount of \( \mathrm{H}_2 \mathrm{S} \) produced. Such understanding helps in setting up stoichiometric calculations and solving for the unknowns in a problem.
Analyzing these chemical reactions provides insights into the nature of reactants and products. It also highlights the importance of balancing equations as it ensures the conservation of mass.

Percent Composition

Percent composition is a valuable calculation in chemistry that shows the percentage by mass of each element in a compound. More broadly, it can indicate the proportion of different substances in a mixture, as was done in the exercise.

For our problem, you first calculate the masses of \( \mathrm{ZnS} \) and \( \mathrm{PbS} \) using the amount in moles obtained previously and their respective molar masses. Then, using the overall sample mass, you determine the mass percentage of each compound using the formula:

• \( \% \ \text{compound} = \left( \frac{\text{mass of compound}}{\text{total mass of mixture}} \right) \times 100 \)

This allows us to calculate the percent composition by taking the mass of each sulfide and dividing by the total mass, then multiplying by 100 to get a percentage. In our exercise, the mixture contained 47.34% \( \mathrm{ZnS} \) and 52.66% \( \mathrm{PbS} \).
Understanding percent composition not only helps in quantifying contributions of components in a mixture, but also in predicting reaction yields and purities.