Problem 39 - General Chemistry

Question

Zinc metal reacts with yellow crystals of sulfur in a fiery reaction to produce a white powder of zinc sulfide. A chemist determines that \(65.4 \mathrm{~g}\) of zinc reacts with \(32.1 \mathrm{~g}\) of sulfur. How many grams of zinc sulfide could be produced from \(36.9 \mathrm{~g}\) of zinc metal?

Step-by-Step Solution

Answer

55.0 grams of zinc sulfide can be produced.

Step 1: Write the chemical reaction

Zinc reacts with sulfur to form zinc sulfide. The chemical equation for this reaction is \( \text{Zn} + \text{S} \rightarrow \text{ZnS} \). This indicates that one mole of zinc reacts with one mole of sulfur to produce one mole of zinc sulfide.

Step 2: Use molar masses to find the stoichiometry

The molar mass of zinc (Zn) is approximately 65.4 g/mol and that of sulfur (S) is about 32.1 g/mol. Zinc sulfide (ZnS) has a molar mass of approximately 97.5 g/mol (65.4 g/mol for Zn + 32.1 g/mol for S). The stoichiometry from the equation confirms a 1:1:1 molar ratio.

Step 3: Calculate the theoretical yield from given zinc

From the reaction, we know that 65.4 g of zinc reacts completely with 32.1 g of sulfur to produce 97.5 g of zinc sulfide. We need to calculate how much zinc sulfide can be produced from 36.9 g of zinc. We set up a proportion: \( \frac{97.5 \, \text{g ZnS}}{65.4 \, \text{g Zn}} = \frac{x}{36.9 \, \text{g Zn}} \). Solving for \( x \) gives us \( x = \frac{97.5 \, \times \, 36.9}{65.4} \).

Step 4: Perform the calculation

Now, calculate: \( x = \frac{97.5 \, \times \, 36.9}{65.4} \approx 55.0 \, \text{g} \). Therefore, 36.9 g of zinc would produce approximately 55.0 g of zinc sulfide.

Key Concepts

Chemical Reactions

Understanding chemical reactions is essential in stoichiometry, especially for reactions like zinc and sulfur forming zinc sulfide. A chemical reaction involves the transformation of reactants into products, signified by a balanced chemical equation. Here, the reaction between zinc (Zn) and sulfur (S) forms zinc sulfide (ZnS). This can be seen in the equation: \(\text{Zn} + \text{S} \rightarrow \text{ZnS}\).

• The reactants are zinc and sulfur, which are on the left side of the equation.
• The product, zinc sulfide, is on the right side.
• Each element must balance on both sides, ensuring mass conservation during the reaction.

Understanding this balance allows chemists to predict how much product forms from given reactants. This balance is foundational to understanding the stoichiometric calculations that follow.

Molar Mass

Molar mass is a vital concept for calculating the amounts involved in chemical reactions. It represents the mass of one mole of a substance, expressed in grams per mole (g/mol). In the exercise, understanding molar masses helps determine how much zinc sulfide can be produced from a given amount of zinc.

Let's look at the molar masses involved:

• Zinc (Zn) has a molar mass of approximately \(65.4\, \text{g/mol}\).
• Sulfur (S) has a molar mass of about \(32.1\, \text{g/mol}\).
• Zinc sulfide (ZnS), when combined, has a molar mass of \(97.5\, \text{g/mol}\) (calculated as \(65.4 + 32.1\)).

These values help establish the relationship between the quantity of zinc and the zinc sulfide it produces. By comparing given weights to their molar masses, you can predict the amount of product formed in a chemical reaction, essential for determining the theoretical yield.

Theoretical Yield

The theoretical yield in a chemical reaction is the amount of product that could be formed if the reactants react completely under ideal conditions. It represents the maximum quantity predicted by stoichiometric calculations based on reactant amounts. In the given zinc and sulfur reaction, calculating the theoretical yield involves using the stoichiometry of the reaction and the molar masses of the substances involved.

The process is as follows:

• You've learned that \(65.4\,\text{g}\) of zinc can yield \(97.5\,\text{g}\) of zinc sulfide.
• To find the theoretical yield from \(36.9\,\text{g}\) of zinc, use a proportion: \(\frac{97.5 \, \text{g ZnS}}{65.4 \, \text{g Zn}} = \frac{x}{36.9 \, \text{g Zn}}\).
• Solving for \(x\), representing the yield from \(36.9\,\text{g}\), involves simple algebraic manipulation.

This step lets you predict that \(36.9\,\text{g}\) of zinc potentially produces about \(55.0\,\text{g}\) of zinc sulfide under ideal circumstances. Establishing such yields is essential in chemistry for planning and executing experiments efficiently.