Question

Complete the following table for the reaction of cobalt and sulfur and calculate the moles of \(\mathrm{Co}, \mathrm{S},\) and \(\mathrm{Co}_{2} \mathrm{~S}_{3}\) after reac tion according to the balanced equation: $$2 \mathrm{Co}(s)+3 \mathrm{~S}(s) \longrightarrow \mathrm{Co}_{2} \mathrm{~S}_{3}(s)$$ $$\begin{array}{llll}\hline \text { Experiment } & \text { mol Co } & \text { mol S } & \text { mol }\mathrm{Co}_{2} \mathrm{~S}_{3} \\\\\hline \begin{array}{l}\text { (a) before reaction: } \\\\\text { after reaction: }\end{array} & 1.00 & 1.00 & 0.00 \\\\\text { (b) before reaction: } & 2.00 & 3.00 & 0.00 \\\\\text { after reaction: } & & & \\\\\hline\end{array}$$

Short answer

After part (a) reaction: 0.33 mol \(\mathrm{Co}_{2}\mathrm{~S}_{3}\), 0.33 mol Co remain. After part (b) reaction: 1.00 mol \(\mathrm{Co}_{2}\mathrm{~S}_{3}\), 0.00 mol Co and S remain.

Step-by-step solution

Balanced Equation Analysis

First, analyze the balanced equation: \(2 \mathrm{Co}(s) + 3 \mathrm{S}(s) \longrightarrow \mathrm{Co}_{2}\mathrm{~S}_{3}(s)\). This implies that 2 moles of Co react with 3 moles of S to produce 1 mole of \(\mathrm{Co}_{2}\mathrm{~S}_{3}\).

Calculate Reaction Limiting Reactant for Part (a)

For part (a), we start with 1.00 mol of Co and 1.00 mol of S. From the balanced equation, 1.00 mol of Co would require \(1.5\) mol of S to fully react, but only 1.00 mol of S is available. Thus, S is the limiting reactant.

Determine Moles of \(\mathrm{Co}_{2}\mathrm{~S}_{3}\) Formed in Part (a)

Since S is limiting with 1.00 mol available, and \(3\) moles of S will produce \(1\) mole of \(\mathrm{Co}_{2}\mathrm{~S}_{3}\), the number of moles of \(\mathrm{Co}_{2}\mathrm{~S}_{3}\) formed is \(\frac{1.00}{3} = 0.33\) mol.

Calculate Unreacted Co in Part (a)

From the balanced equation, \(2\) mol of Co is required for \(3\) mol of S. Hence \(1\) mol of S would use \(\frac{2}{3} \times 1.00 = 0.67\) mol of Co. Thus, remaining Co after reaction = \(1.00 - 0.67 = 0.33\) mol.

Calculate Moles After Reaction for Part (b)

Before reaction in part (b), we have 2.00 mol Co and 3.00 mol S. According to the balanced equation, these will react to form \(1\) mole of \(\mathrm{Co}_{2}\mathrm{~S}_{3}\) (since \(2 \mathrm{Co}\) and \(3 \mathrm{S}\) fully react). Thus, moles after reaction: \(\mathrm{Co} = 0.00\), \(\mathrm{S} = 0.00\), and \(\mathrm{Co}_{2}\mathrm{~S}_{3} = 1.00\).

Key concepts

Chemical Reactions

Chemical reactions involve breaking and forming bonds between atoms, leading to new substances. In the given reaction, cobalt (Co) and sulfur (S) combine to form cobalt sulfide (\(\mathrm{Co}_{2}\mathrm{~S}_{3}\)).
A chemical reaction is symbolized by a chemical equation. This illustrates the reactants (cobalt and sulfur in this instance) and the resultant products (cobalt sulfide).
The arrow indicates the direction of the reaction, showing how the reactants transform.

• Cobalt and sulfur react by trading electrons to form bonds, creating a new compound, cobalt sulfide.
• Each atom involved in the reaction follows the conservation of mass principle, meaning no atom is lost or gained.
• Throughout the reaction, elements in the reactants recombine to form different configurations, leading to a unique outcome.

Understanding these changes lets us predict the materials involved before and after the reaction. This is crucial for calculating amounts needed and products formed.

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that runs out first, thereby determining the maximum amount of product that can be generated.
For parts (a) and (b) in our example:

• Step 2 showed that in part (a), we had 1.00 mol of Co and 1.00 mol of S. According to the equation, 1 mole of Co requires 1.5 moles of S to fully react. We only have 1 mole of sulfur, making sulfur the limiting reactant.
• For part (b), however, 2 moles of Co react with 3 moles of S completely, leaving no excess of either reactant.

Being able to identify the limiting reactant is essential because it allows us to calculate how much of the product can be formed. It's like cooking: if you want to bake cookies, but run out of flour or sugar first, the ingredient you run out of limits how many cookies you make. Knowing the limiting reactant helps allocate resources efficiently and predict or control the output of a chemical process.

Balanced Chemical Equation

A balanced chemical equation respects the law of conservation of mass. It ensures that the number of atoms for each element is equal on both sides of the equation.
For the cobalt and sulfur reaction, the balanced equation is:
\[2 \mathrm{Co}(s) + 3 \mathrm{~S}(s) \longrightarrow \mathrm{Co}_{2}\mathrm{~S}_{3}(s)\]

• The equation shows that 2 atoms of cobalt combine with 3 atoms of sulfur to form one new substance.
• Balancing ensures that both sides of the equation have the same number of each type of atom, a key to reflecting physical reality.
• Each coefficient in a balanced equation tells us the proportionate amounts of reactants used and products formed.

Balancing equations is crucial in understanding and predicting the outcomes of reactions. By doing so, chemists can calculate exactly how much of a chemical is necessary or how much can be produced.