Question

The chief ore of \(\mathrm{Zn}\) is the sulphide, \(\mathrm{ZnS}\). The ore is concentrated by froth floatation process and then heated in air to convert \(\mathrm{ZnS}\) to \(\mathrm{ZnO}\). $$ \begin{aligned} &2 \mathrm{ZnS}+3 \mathrm{O}_{2} \stackrel{80 \%}{\longrightarrow} 2 \mathrm{ZnO}+2 \mathrm{SO}_{2} \\ &\mathrm{ZnO}+\mathrm{H}_{2} \mathrm{SO}_{4} \stackrel{100 \%}{\longrightarrow} \mathrm{ZnSO}_{4}+\mathrm{H}_{2} \mathrm{O} \\ &2 \mathrm{ZnSO}_{4}+2 \mathrm{H}_{2} \mathrm{O} \stackrel{80 \%}{\longrightarrow} 2 \mathrm{Zn}+2 \mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{O}_{2} \end{aligned} $$ The number of moles of \(\mathrm{ZnS}\) required for producing 2 moles of \(\mathrm{Zn}\) will be : (a) \(3.125\) (b) 2 (c) \(2.125\) (d) 4

Short answer

3.125

Step-by-step solution

Analyze the First Reaction

Consider the first reaction: \(2 \mathrm{ZnS} + 3 \mathrm{O}_{2} \rightarrow 2 \mathrm{ZnO} + 2 \mathrm{SO}_{2}\). Here, 2 moles of \(\mathrm{ZnS}\) are required to produce 2 moles of \(\mathrm{ZnO}\). Since the demand is to produce 2 moles of \(\mathrm{Zn}\) eventually, and 2 moles of \(\mathrm{ZnO}\) will lead to 2 moles of \(\mathrm{Zn}\) in subsequent reactions, we need 2 moles of \(\mathrm{ZnS}\) in this step.

Account for Process Efficiency

The reaction efficiency for the first reaction is given as 80%. This means that to actually get the 2 moles of \(\mathrm{ZnO}\), we need to start with more than 2 moles of \(\mathrm{ZnS}\). To account for this, divide the required moles of \(\mathrm{ZnS}\) by the efficiency: \(\frac{2 \text{ moles }}{0.8} = 2.5 \text{ moles of } \mathrm{ZnS}\).

Assess Overall Yield

To ensure that the complete process from \(\mathrm{ZnS}\) to \(\mathrm{Zn}\) yields 2 moles of \(\mathrm{Zn}\) at the end, we need to make sure that losses in the first step are accounted for across all steps of the process. Since no other steps reduce the yield of \(\mathrm{ZnO}\) to \(\mathrm{Zn}\), and the last step also has an 80% efficiency, this does not alter the required amount of \(\mathrm{ZnS}\). Thus, the required moles of \(\mathrm{ZnS}\) remains at 2.5.

Key concepts

Chemical Reactions

Understanding chemical reactionsis fundamental to solving stoichiometry problems. A chemical reaction is a process where reactants are transformed into products through the breaking and forming of chemical bonds. In our exercise, we see a sequence of reactions that transform \textbf{ZnS} (zinc sulfide) into \textbf{Zn} (zinc metal), which involves intermediate steps producing \textbf{ZnO} (zinc oxide) and other compounds.
To tackle such problems, you need to balance the chemical equation, ensuring that the same number of each type of atom appears on both sides of the reaction. This balancing act respects the law of conservation of mass. For example, in the first provided reaction, \(2 \text{ZnS} + 3 \text{O}_{2} \rightarrow 2 \text{ZnO} + 2 \text{SO}_{2}\), two molecules of zinc sulfide react with three molecules of oxygen to form two molecules each of zinc oxide and sulfur dioxide.
Balancing equations allows us to use stoichiometry to determine the amount of reactants required to produce a certain amount of product under ideal circumstances. However, in real-world scenarios, reactions often do not proceed to completion or have side reactions, which is why reaction efficiency must also be taken into account.

Mole Concept

The mole concept is a cornerstone of chemistry that provides a way to quantify substances based on how many particles, atoms, ions, or molecules they contain. One mole is equivalent to Avogadro's number (\(6.02 \times 10^{23}\) particles). Stoichiometry heavily relies on this concept because it allows the translation of mass into moles, which is indispensable when using balanced equations to relate quantities of reactants and products.
In the context of our exercise, to determine how many moles of \textbf{ZnS} are required to produce 2 moles of \textbf{Zn}, we must first understand the molar ratios between reactants and products in each step. Here’s where stoichiometry bridges the mole concept with chemical reactions. We start by looking at the balanced equations, which, in this case, indicate a 1:1 molar ratio between \textbf{ZnO} and \textbf{Zn} in the final step of the reaction sequence.
By using this ratio, we can map a direct relationship from \textbf{ZnS} to \textbf{Zn}, ensuring that each step's stoichiometry is respected. However, we must remember that the mole concept doesn't take into account real-world inefficiencies, hence the need to adjust calculations to include reaction efficiency as well.

Reaction Efficiency

When solving stoichiometry problems, accounting for reaction efficiency is necessary to achieve realistic results. Reaction efficiency, or yield, refers to the percentage of the theoretical amount of product that is actually produced when a chemical reaction is carried out in practice.
For example, the first step in our exercise has an 80% efficiency, meaning that only 80% of the theoretical yield of \textbf{ZnO} is obtained from \textbf{ZnS}. To accommodate for the losses due to inefficiency, calculations must be adjusted accordingly. As seen in the solution, for a theoretical yield of 2 moles of \textbf{ZnO}, we need to start with \(\frac{2 \text{ moles of ZnS}}{0.8}\), which gives us 2.5 moles of \textbf{ZnS}.
It's also important to follow this adjustment through the entire series of reactions when necessary. If subsequent reactions show different efficiencies, we would need to make additional adjustments. In our case, the second reaction is 100% efficient, which doesn’t require any adjustment, but for the last step, which is also 80% efficient, it's already accounted for in the initial adjustment. Making thorough and accurate calculations like this is crucial, as it helps ensure enough starting material is used to end up with the desired amount of final product.