Question

A 0.696 mol sample of \(\mathrm{Cu}\) is added to \(136 \mathrm{mL}\) of \(6.0 \mathrm{M}\) HNO \(_{3}\). Assuming the following reaction is the only one that occurs, will the Cu react completely? $$\begin{aligned} 3 \mathrm{Cu}(\mathrm{s})+8 \mathrm{HNO}_{3}(\mathrm{aq}) & \longrightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) +4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+2 \mathrm{NO}(\mathrm{g}) \end{aligned}$$

Short answer

No, the copper will not react completely with the nitric acid due to the insufficient amount of nitric acid.

Step-by-step solution

Calculate the amount of HNO3 in moles

First, the moles of HNO3 are calculated from the given volume and molarity, using the equation \( Molarity = Moles/Volume \). Here the volume should be converted to liters by dividing by 1000, so the calculation would be: \( Moles_{HNO3} = Molarity_{HNO3} * Volume_{HNO3}/1000 = 6M * 136ml/1000 = 0.816 mol \)

Determine the limiting reactant

Next, it's necessary to determine which reactant is limiting, i.e., will run out first and stop the reaction. According to the balanced equation, 3 moles of copper are needed for 8 moles of nitric acid. But there are 0.696 mol of copper and 0.816 mol of nitric acid in the reaction. So, calculate how much nitric acid is needed for available copper and how much copper is needed for available nitric acid. This is found from the following: \( Needed HNO3 = 8/3 * Available Cu = 8/3 * 0.696 = 1.85 mol \). And, \( Needed Cu = 3/8 * Available HNO3 = 3/8 * 0.816 = 0.306 mol \). Comparing these, we see that the amount of needed nitric acid for the available copper is more than the provided, so nitric acid is the limiting reactant.

Determine if all copper reacts

For the last step, determine if all copper reacts. From the calculations, the available nitric acid required 0.306 mol of copper for the reaction to complete. The total amount of copper available was 0.696 mol, which is much more than needed. Therefore, not all the copper will react completely.

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that involves calculating the amounts of reactants and products in a chemical reaction. In any given reaction, stoichiometry helps us to precisely know how much of each reactant is needed to fully react and form products. This involves converting the balanced chemical equation into a quantitative view of the substances involved.

In our copper and nitric acid reaction example, the stoichiometry is derived from the balanced equation:

• 3 moles of copper (\(\text{Cu}\)) react with 8 moles of nitric acid (\(\text{HNO}_3\)).
• As a result, 3 moles of copper (II) nitrate, 4 moles of water, and 2 moles of nitrogen monoxide are formed.

By understanding this stoichiometric relationship, you can calculate the exact quantities needed for each reactant or product. It is crucial for determining the limiting reactant, which eventually defines the extent of the reaction.

Molarity

Molarity is a way to express the concentration of a solution, measured in moles of solute per liter of solution. The formula to determine molarity (\(M\)) is:

• \(M = \text{moles of solute} / \text{liters of solution}\).

In the problem, the molarity of nitric acid is given as 6.0 M, and you have 136 mL of this solution. To convert the volume to liters, divide by 1000:

• \(136\, \text{mL} = 0.136\, \text{L}\).

Now, you can find the moles of \(\text{HNO}_3\):

• \(\text{moles} = 6.0\, M \times 0.136\, L = 0.816\, \text{mol}\).

Molarity calculations like these are essential to understanding how much reactant is available to contribute to the reaction.

Chemical Reactions

Chemical reactions describe how substances interact and transform into new substances. They are defined by a balanced chemical equation, which ensures that the same number of atoms for each element are present on both the reactant and product sides.

In our context, the reaction between copper and nitric acid illustrates the conversion of solid copper into soluble copper nitrate, alongside the development of water and nitrogen monoxide gas. The balanced equation is:

• \(3\, \text{Cu} (\text{s}) + 8\, \text{HNO}_3 (\text{aq}) \rightarrow 3\, \text{Cu(NO}_3\text{)}_2 (\text{aq}) + 4\, \text{H}_2\text{O} (\text{l}) + 2\, \text{NO} (\text{g})\)

This equation not only tells us what products form, but also assures that no atoms are lost and energy is conserved. It provides a blueprint for practical calculations, such as determining limiting reactants and yields.

Copper and Nitric Acid Reaction

When copper reacts with nitric acid, it's a fascinating redox reaction, as copper is oxidized and nitric acid is reduced. This is a detailed process where electrons are exchanged between substances, leading to changes in their oxidation states.

In detail:

• Copper acts as the reducing agent, losing electrons and turning into copper ions.
• Nitric acid is reduced, oxidizing nitrogen to form nitrogen monoxide gas.

In the given problem, understanding this reaction is pivotal to identifying copper's fate in the process. Since we found nitric acid to be the limiting reactant, it ultimately determines that not all of the 0.696 mol of copper reacts completely. The leftover copper indicates that the availability of nitric acid constrained the reaction, meaning it couldn't continue past a certain point.