The first step in the industrial recovery of zinc from the zinc sulfide ore is roasting, that is, the conversion of \(\mathrm{ZnS}\) to \(\mathrm{ZnO}\) by heating: $$\begin{array}{r}2 \mathrm{ZnS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{ZnO}(s)+2 \mathrm{SO}_{2}(g) \\ \Delta H=-879 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ Calculate the heat evolved (in kJ) per gram of ZnS roasted.

Short Answer

Expert verified
The heat evolved per gram of ZnS roasted is 4.51 kJ/g.

Step by step solution

01

Understand the reaction

The balanced chemical equation is \(2 \mathrm{ZnS}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{ZnO}(s)+2 \mathrm{SO}_{2}(g)\), where \(\Delta H=-879 \mathrm{~kJ} / \mathrm{mol}\). This means that the reaction of 2 mol of ZnS with 3 mol O2 produces 2 mol of ZnO and 2 mol of SO2, releasing 879 kJ of heat.
02

Calculate the molar mass of ZnS

The molar mass of ZnS (Zinc Sulfide) can be calculated by adding the atomic mass of Zinc (Zn, 65.38 g/mol) and Sulfur (S, 32.06 g/mol). The molar mass of ZnS is \(65.38 + 32.06 = 97.44 \, g/mol\).
03

Calculate the heat evolved per mol of ZnS

According to the balanced equation, 2 mol of ZnS reacts to give off 879 kJ of heat. So, the heat evolved per mol of ZnS would be \(879/2 = 439.5 \, kJ/mol\).
04

Calculate the heat evolved per gram of ZnS

To find out the heat evolved per gram, we need to convert the energy per mol to energy per gram. This can be done by dividing the energy per mol by the molar mass of ZnS. So, the heat evolved per gram of ZnS is \(439.5 \, kJ/mol / 97.44 \, g/mol = 4.51 \, kJ/g\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reaction Stoichiometry
Chemical reaction stoichiometry is a branch of chemical science that involves the quantitative relationships between the reactants and products in a chemical reaction. Understanding stoichiometry is essential for predicting how much product will form from a given amount of reactants, which is crucial in industrial processes such as the roasting of zinc sulfide ore to produce zinc oxide.

In the example of the zinc sulfide ore roasting, the balanced chemical equation $$2 \mathrm{ZnS}(s) + 3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{ZnO}(s) + 2 \mathrm{SO}_{2}(g)$$illustrates the stoichiometric relationship. The coefficients in front of each substance indicate their respective molar ratios. For every 2 moles of zinc sulfide (\(\mathrm{ZnS}\)), 3 moles of oxygen gas (\(\mathrm{O}_{2}\)) are required, and the reaction will yield 2 moles of zinc oxide (\(\mathrm{ZnO}\)) and 2 moles of sulfur dioxide (\(\mathrm{SO}_{2}\)).

Understanding the stoichiometry is not just about balancing chemical equations; it's also critical in determining the efficiency and yield of the industrial process. For instance, if you start with an excess of \(\mathrm{ZnS}\), not all of it will be converted to \(\mathrm{ZnO}\) unless there is enough \(\mathrm{O}_{2}\) present. Therefore, accurate calculation and measurement are key to ensuring the chemical process is both effective and cost-efficient.
Thermochemistry
Thermochemistry is the study of the heat energy associated with chemical reactions and physical transformations. The change in enthalpy (\(\Delta H\)) of a reaction, which is the heat absorbed or released under constant pressure, is a fundamental concept in this field.

In the context of our roasting reaction of zinc sulfide, the given enthalpy change is \(-879 \mathrm{~kJ/mol}\), indicating that the reaction is exothermic; it releases heat to the surroundings. This amount of energy is substantial when considering the industrial application, as it impacts the energy requirements and overall sustainability of the process. For instance, the heat evolved can be harnessed to power other stages of ore processing, enhancing energy efficiency.

Understanding the enthalpy change is also vital for safety considerations in industrial settings where controlling the temperature is crucial to prevent accidents. It can dictate the need for cooling systems to manage excess heat or insulation strategies to maintain optimal temperatures for the reaction to proceed.
Molar Mass Calculation
Molar mass calculation is a process used to determine the mass of one mole of a substance, expressed typically in grams per mole (\(g/mol\)). It is an essential aspect of chemical stoichiometry, as it allows chemists to convert between the mass of a substance and the amount in moles.

For example, the molar mass of zinc sulfide (\(\mathrm{ZnS}\)) is found by adding the atomic masses of zinc (\(\mathrm{Zn}\), 65.38 \(g/mol\)) and sulfur (\(\mathrm{S}\), 32.06 \(g/mol\)). Therefore, the molar mass of \(\mathrm{ZnS}\) is \(65.38 + 32.06 = 97.44 \, g/mol\).

When calculating the heat evolved per gram of \(\mathrm{ZnS}\) roasted, we use this molar mass in conjunction with the thermochemical data to find that \(4.51 \, kJ/g\) of energy is released. This information is critical for adjusting the scale of the operation to the desired output and for ensuring the correct amount of ore is processed to meet energy requirements. In any industrial process, accurate molar mass calculation underpins the ability to standardize reactions, ensure product quality, and control production costs.

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Most popular questions from this chapter

From the following heats of combustion, $$\begin{array}{r}\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \\\\\Delta H_{\mathrm{rxn}}^{\circ}=-726.4 \mathrm{~kJ} / \mathrm{mol} \\\\\mathrm{C}(\mathrm{graphite})+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \\\\\Delta H_{\mathrm{rxn}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ calculate the enthalpy of formation of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) from its elements: $$\mathrm{C}(\text { graphite })+2 \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow\mathrm{CH}_{3} \mathrm{OH}(l)$$

A driver's manual states that the stopping distance quadruples as the speed doubles; that is, if it takes \(30 \mathrm{ft}\) to stop a car moving at \(25 \mathrm{mph}\) then it would take \(120 \mathrm{ft}\) to stop a car moving at \(50 \mathrm{mph}\). Justify this statement by using mechanics and the first law of thermodynamics. [Assume that when a car is stopped, its kinetic energy \(\left(\frac{1}{2} m u^{2}\right)\) is totally converted to heat.]

A person ate 0.50 pound of cheese (an energy intake of \(4000 \mathrm{~kJ}\) ). Suppose that none of the energy was stored in his body. What mass (in grams) of water would he need to perspire in order to maintain his original temperature? (It takes \(44.0 \mathrm{~kJ}\) to vaporize 1 mole of water.

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) have the same empirical formula. In fact, benzene can be made from acetylene as follows: $$ 3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l) $$ The enthalpies of combustion for \(\mathrm{C}_{2} \mathrm{H}_{2}\) and \(\mathrm{C}_{6} \mathrm{H}_{6}\) are \(-1299.4 \mathrm{~kJ} / \mathrm{mol}\) and \(-3267.4 \mathrm{~kJ} / \mathrm{mol},\) respectively. Calculate the standard enthalpies of formation of \(\mathrm{C}_{2} \mathrm{H}_{2}\) and \(\mathrm{C}_{6} \mathrm{H}_{6}\) and hence the enthalpy change for the formation of \(\mathrm{C}_{6} \mathrm{H}_{6}\) from \(\mathrm{C}_{2} \mathrm{H}_{2}\).

A gas expands and does \(P-V\) work on the surroundings equal to \(325 \mathrm{~J}\). At the same time, it absorbs \(127 \mathrm{~J}\) of heat from the surroundings. Calculate the change in energy of the gas.

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