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Chapter 9: Problem 85

When elemental copper is placed in a solution of silver nitrate, the following oxidationreduction reaction takes place, forming clemental silver: $$ \mathrm{Cu}(s)+2 \mathrm{AgNO}_{3}(a q) \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Ag}(s) $$ What mass of copper is required to remove all the silver from a silver nitrate solution containing \(1.95 \mathrm{mg}\) of silver nitrate?

Short Answer

Expert verified
Approximately \(3.65 \times 10^{-4}\) grams of copper is required to remove all the silver from a silver nitrate solution containing 1.95 mg of silver nitrate.

Step by step solution

01

Convert the mass of silver nitrate to moles

To do this, we first need the molar mass of silver nitrate, which can be calculated as follows: \[ M_{\text{AgNO}_3} = M_{\text{Ag}} + M_{\text{N}} + 3\times M_{\text{O}} \] Using the periodic table, we find the molar mass of each element: \(M_{\text{Ag}} = 107.87\,\text{g/mol}\), \(M_{\text{N}} = 14.01\,\text{g/mol}\), \(M_{\text{O}} = 16.00\,\text{g/mol}\). Plugging in the values: \[ M_{\text{AgNO}_3} = 107.87 + 14.01 + 3\times 16.00 = 169.88\,\text{g/mol} \] Now, we can find the moles of silver nitrate using its mass and molar mass: \[ \text{moles}_{\text{AgNO}_3}\! = \frac{\text{mass}_{\text{AgNO}_3}}{M_{\text{AgNO}_3}}\! = \frac{1.95\,\text{mg}}{169.88\,\text{g/mol}} = \frac{0.00195\,\text{g}}{169.88\,\text{g/mol}}\! \approx 1.148 \times 10^{-5}\,\text{mol} \]
02

Find the moles of elemental copper based on the stoichiometry of the reaction

From the balanced chemical equation, we can see that 1 mole of elemental copper (Cu) reacts with 2 moles of silver nitrate (AgNO3) according to the following reaction: \[ \text{Cu}(s) + 2 \text{AgNO}_{3}(aq) \rightarrow\text{Cu}(\text{NO}_{3})_{2}(aq) + 2 \text{Ag}(s) \] Thus, the stoichiometric ratio between Cu and AgNO3 is 1:2. So, for every 2 moles of AgNO3, 1 mole of Cu is consumed. We now convert the moles of silver nitrate into moles of elemental copper using the stoichiometric ratio: \[ \text{moles}_{\text{Cu}}\! = \text{moles}_{\text{AgNO}_3}\! \times \frac{1\,\text{mol}(\text{Cu})}{2\,\text{mol}(\text{AgNO}_3)}\! = 1.148 \times 10^{-5}\,\text{mol}\left(\text{AgNO}_3\right)\! \times \frac{1}{2} = 5.74 \times 10^{-6}\,\text{mol} \]
03

Convert the moles of elemental copper to mass

Finally, we can calculate the mass of Cu by multiplying the moles of Cu by its molar mass. From the periodic table, we find that the molar mass of Cu is \(63.55\,\text{g/mol}\). Thus: \[ \text{mass}_{\text{Cu}}\! = \text{moles}_{\text{Cu}}\! \times M_{\text{Cu}}\! = 5.74 \times 10^{-6}\,\text{mol}\times 63.55\,\text{g/mol} \approx 3.65 \times 10^{-4}\,\text{g} \] So, approximately \(3.65 \times 10^{-4}\) grams of copper is required to remove all the silver from a silver nitrate solution containing 1.95 mg of silver nitrate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Weight Calculation
To find out how much copper we need in a chemical reaction, we must first calculate the molecular weight of the substances involved. The molecular weight (or molar mass) is the sum of the atomic masses of all the atoms in a molecule. Here's an example given in the exercise: silver nitrate (AgNO鈧�).

To calculate the molecular weight of AgNO鈧�:
  • Find the atomic weights of silver (Ag), nitrogen (N), and oxygen (O) using the periodic table.
  • Ag = 107.87 g/mol, N = 14.01 g/mol, and O = 16.00 g/mol.
  • Calculate the molecular weight: \[ M_{\text{AgNO}_3} = M_{\text{Ag}} + M_{\text{N}} + 3 \times M_{\text{O}} = 107.87 + 14.01 + 3\times 16.00 = 169.88 \, \text{g/mol} \]
Calculating the molecular weight helps us convert between moles and grams, crucial for determining quantities in reactions. Understanding these conversions is foundational in stoichiometry.
Oxidation-Reduction Reaction
In chemical processes, oxidation-reduction (redox) reactions are vital as they involve the transfer of electrons. The given exercise showcases a redox reaction between copper and silver nitrate:

\[ \text{Cu}(s) + 2 \text{AgNO}_3(aq) \rightarrow \text{Cu(NO}_3)_2(aq) + 2 \text{Ag}(s) \]

In this equation:
  • Copper (Cu) loses electrons to become Cu(NO鈧�)鈧�, undergoing oxidation.
  • Silver ions (Ag鈦�) gain electrons to form elemental silver (Ag), undergoing reduction.
This type of reaction is pivotal because it allows for the formation of new compounds and elements. Understanding the terms "oxidation" (loss of electrons) and "reduction" (gain of electrons) is crucial. Identifying which substance is oxidized and which is reduced lets us predict the outcome of the reaction.
Chemical Reaction Balancing
Balancing a chemical reaction ensures the conservation of mass. Each type of atom should have the same quantity on both sides of the equation. In our example, the balanced equation is:

\[ \text{Cu}(s) + 2 \text{AgNO}_3(aq) \rightarrow \text{Cu(NO}_3)_2(aq) + 2 \text{Ag}(s) \]

In this balanced equation:
  • One atom of copper reacts, producing one copper nitrate molecule, ensuring copper is balanced.
  • Two silver ions react, forming two silver atoms, balancing the silver.
  • The two NO鈧� groups involved on both sides keep nitrate balanced.
Balancing equations is fundamental to ensure that the same amounts of matter exist before and after the reaction. This principle is based on the law of conservation of mass, a core idea in chemistry. By balancing reactions, we can correctly predict the quantities of reactants needed and products formed.

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Most popular questions from this chapter

If common sugars are heated too strongly, they char as they decompose into carbon and water vapor. For example, if sucrose (table sugar) is heated, the reaction is $$ \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s) \rightarrow 12 \mathrm{C}(s)+11 \mathrm{H}_{2} \mathrm{O}(g) $$ What mass of carbon is produced if 1.19 g of sucrose decomposes completely?

The traditional method of analysis for the amount of chloride ion prescnt in a sample is to dissolve the sample in water and then slowly to add a solution of silver nitrate. Silver chloride is very insoluble in water, and by adding a slight excess of silver nitrate, it is possible to effectively remove all chloride ion from the sample. Suppose a \(1.054-g\) sample is known to contain \(10.3 \%\) chloride ion by mass. What mass of silver nitrate must be used to completely precipitate the chloride ion from the sample? What mass of silver chloride will be obtained?

Consider the balanced equation $$ \mathrm{C}_{3} \mathrm{H}_{3}(g)+5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ What mole ratio enables you to calculate the number of moles of oxygen needed to react exactly with a given number of molcs of \(\mathrm{C}_{3} \mathrm{H}_{3}(g)\) ? What mole ratios enable you to calculate how many moles of each product form from a given number of moles of \(\mathrm{C}_{3} \mathrm{H}_{8}\) ?

Consider the following unbalanced chemical equation. $$ \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$ If \(67.4 \mathrm{~g}\) of lithium hydroxide reacts with excess carbon dioxide, what mass of lithium carbonate will be produced?

For each of the following unbalanced chemical equations, suppose that exactly \(1.00 \mathrm{~g}\) of each reactant is taken. Determine which reactant is limiting, and calculate what mass of the product in boldface is expected (assuming that the limiting reactant is completely consumed). a. \(\mathrm{CS}_{2}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{SO}_{2}(g)\) b. \(\mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{CN}_{2} \mathrm{H}_{4} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{MnO}_{2}(s) \rightarrow \mathrm{MnO}(s)+\mathbf{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{I}_{2}(l)+\mathrm{Cl}_{2}(g) \rightarrow \mathbf{I C l}(g)\)
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