Question

Copper metal can reduce silver ions to metallic silver. The copper is oxidized to copper ions according to the reaction $$ 2 \mathrm{Ag}^{+}(a q)+\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s) $$ A copper strip with a mass of \(2.00 \mathrm{~g}\) is dipped into a solution of \(\mathrm{AgNO}_{3} .\) After some time has elapsed, the copper strip is coated with silver. The strip is removed from the solution, dried, and weighed. The coated strip has a mass of \(4.18 \mathrm{~g}\). What are the masses of copper and silver metals in the strip? (Hint: Remember that the copper metal is being used up as silver metal forms.)

Short answer

Question: Calculate the masses of copper and silver metals in a strip after a chemical reaction. The initial mass of the copper strip is 2.00 g and the mass of the coated strip after the reaction is 4.18 g. Answer: After the chemical reaction, the mass of copper in the strip is 1.36 g and the mass of silver is 2.18 g.

Step-by-step solution

Calculate moles of copper initially present

As we know the mass of copper strip initially, we can find the moles of copper using the formula: moles of copper = (mass of copper) / (molar mass of copper) The mass of copper = 2.00 g The molar mass of copper = 63.55 g/mol (from periodic table) So, moles of copper = 2.00 g / 63.55 g/mol = 0.0315 mol

Determine the mass of silver deposited on the copper strip

We know that the coated strip has a mass of 4.18 g and the initial mass of copper strip was 2.00 g. Thus, we can calculate the mass of silver deposited by subtracting the initial mass of copper strip from the mass of the coated strip: Mass of silver deposited = Mass of coated strip - Initial mass of copper strip Mass of silver deposited = 4.18 g - 2.00 g = 2.18 g

Calculate moles of silver deposited

Now, we can find the moles of silver using the formula: moles of silver = (mass of silver) / (molar mass of silver) The mass of silver = 2.18 g The molar mass of silver = 107.87 g/mol (from periodic table) So, moles of silver = 2.18 g / 107.87 g/mol = 0.0202 mol

Determine the moles of copper used in the reaction

Based on the balanced chemical reaction, the mole ratio between silver ions and copper is 2:1. Therefore, for every 2 moles of silver ions, 1 mole of copper is used. So, moles of copper used = moles of silver deposited / 2 moles of copper used = 0.0202 mol / 2 = 0.0101 mol

Calculate the new moles of copper in the strip

Now, we can find the new moles of copper present in the strip after the reaction by subtracting the moles of copper used from the initial moles of copper: New moles of copper = Initial moles of copper - moles of copper used New moles of copper = 0.0315 mol - 0.0101 mol = 0.0214 mol

Find the final masses of copper and silver in the strip

Finally, we can find the new mass of copper present in the strip by multiplying the new moles of copper by the molar mass of copper: Mass of copper = New moles of copper * Molar mass of copper Mass of copper = 0.0214 mol * 63.55 g/mol = 1.36 g Now we have the masses of copper and silver metals in the strip: Mass of copper = 1.36 g Mass of silver = 2.18 g

Key concepts

Stoichiometry

Stoichiometry is the heart of chemistry that involves the quantitative relationships between the reactants and products in a chemical reaction. By knowing how to apply stoichiometry, students can predict the amounts of substances consumed and produced in a given reaction. The stoichiometric coefficients in a balanced chemical equation tell us the ratio in which substances react and are generated.

For instance, in our exercise, the balanced equation shows that one copper atom reacts with two silver ions. This 1:2 ratio is a stoichiometric relationship that allows us to calculate the amount of copper that reacted given the amount of silver that was deposited. Always ensure the chemical equation is balanced before proceeding with any stoichiometric calculations to get accurate results.

Molar Mass

The concept of molar mass is crucial for converting between the mass of a substance and the moles, which is the unit for the amount of substance in chemistry. It's the mass of one mole of a substance, normally expressed in grams per mole (g/mol). Each element has a unique molar mass based on its atomic weight, which you can find on the periodic table.

In our exercise, we've used the molar masses of copper (63.55 g/mol) and silver (107.87 g/mol) to calculate how many moles of each metal are involved in the reaction. This conversion is fundamental for solving the problem since the chemical equation and stoichiometry require the use of moles rather than grams.

Chemical Reaction Balancing

Balancing chemical reactions is vital to ensure the law of conservation of mass is satisfied; that is, the mass of the reactants equals the mass of the products. Each element must have the same number of atoms on both the reactant and product sides of the equation.

In our exercise, the balanced equation shows a 2:1 ratio between silver ions and copper, allowing us to link the amounts of each substance through stoichiometry. Without a balanced equation, our stoichiometric calculations could give incorrect results, leading to a misunderstanding of the reaction's dynamics. Remember, a well-balanced equation is an accurate representation of the chemical reaction, and it's the basis for all further calculations related to the reaction.

Mole Concept

The mole concept is a bridge between the atomic world and the macroscopic world we experience. One mole is Avogadro's number (\(6.022 \times 10^{23}\) entities) of particles, whether they are atoms, molecules, ions, or electrons. It allows chemists to count by weighing, given that counting individual atoms is impractical.

In the provided exercise, we determine that the copper strip initially contains 0.0315 moles of copper. After the reaction, 0.0101 moles of copper are used up, leaving us with 0.0214 moles in the strip. This calculation relies on the mole concept, demonstrating how moles can be used to systematically track the consumption and production of substances in a chemical reaction.