Question

Zinc metal reacts with aqueous silver nitrate to produce silver metal and aqueous zinc nitrate according to the following equation (unbalanced): $$ \mathrm{Zn}(s)+\mathrm{AgNO}_{3}(a q) \longrightarrow \mathrm{Ag}(s)+\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(a q) $$ What mass of silver metal is produced when \(25.00 \mathrm{~g}\) Zn is added to a beaker containing \(105.5 \mathrm{~g} \mathrm{AgNO}_{3}\) dissolved in \(250 \mathrm{~mL}\) of water. Determine the mass amounts of each substance present in the beaker when the reaction is complete.

Short answer

33.47 g of silver is produced, with 4.696 g Zn and 58.82 g Zn(NO3)2 remaining.

Step-by-step solution

Balance the Chemical Equation

The initial unbalanced equation is \( \mathrm{Zn}(s) + \mathrm{AgNO}_{3}(aq) \rightarrow \mathrm{Ag}(s) + \mathrm{Zn}(\mathrm{NO}_{3})_{2}(aq) \). Balance it by adding coefficients:\[ \mathrm{Zn}(s) + 2\mathrm{AgNO}_{3}(aq) \rightarrow 2\mathrm{Ag}(s) + \mathrm{Zn}(\mathrm{NO}_{3})_{2}(aq) \]

Find Molar Masses

Calculate the molar masses for involved components:- \( \mathrm{Zn} \) is 65.38 g/mol.- \( \mathrm{AgNO}_3 \) is 169.87 g/mol.- \( \mathrm{Ag} \) is 107.87 g/mol.- \( \mathrm{Zn(NO}_3)_2 \) is 189.36 g/mol.

Determine Initial Moles

Calculate moles of each reactant:- Moles of \( \mathrm{Zn} = \frac{25.00 \text{ g}}{65.38 \text{ g/mol}} = 0.3823 \text{ mol} \)- Moles of \( \mathrm{AgNO}_3 = \frac{105.5 \text{ g}}{169.87 \text{ g/mol}} = 0.6209 \text{ mol} \)

Identify Limiting Reactant

The balanced equation shows \(1\) mole of \( \mathrm{Zn} \) reacts with \(2\) moles of \( \mathrm{AgNO}_3\). For 0.3823 mol \( \mathrm{Zn} \), 0.7646 mol \( \mathrm{AgNO}_3 \) is needed. Since we only have 0.6209 mol \( \mathrm{AgNO}_3 \), \( \mathrm{AgNO}_3 \) is the limiting reactant.

Calculate Theoretical Yield of Silver

Using limiting reactant \( \mathrm{AgNO}_3 \), calculate moles of \( \mathrm{Ag} \) produced: - From 0.6209 mol \( \mathrm{AgNO}_3 \), 0.31045 mol \( \mathrm{Ag} \) is produced (since \(2\) moles of \( \mathrm{AgNO}_3\) produce \(2\) moles of \( \mathrm{Ag}\)).Calculate mass: \[ \text{Mass of Ag} = 0.31045 \text{ mol} \times 107.87 \text{ g/mol} = 33.47 \text{ g} \]

Determine Remaining Substance Amounts

Calculate remaining amounts:- \( \mathrm{AgNO}_3 \) is consumed entirely as it is the limiting reactant.- \( \mathrm{Zn} \) remaining: initial moles \( - \) reacted moles: \[ 0.3823 \text{ mol} - 0.31045 \text{ mol} = 0.07185 \text{ mol} \] Convert to mass: \[0.07185 \text{ mol} \times 65.38 \text{ g/mol} = 4.696 \text{ g} \]- Mass of \( \mathrm{Zn(NO}_3)_2 \) produced is: \[ 0.31045 \text{ mol} \times 189.36 \text{ g/mol} = 58.82 \text{ g} \]

Key concepts

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, undergo transformations to create new substances, known as products. These reactions represent changes at the molecular level, where bonds are broken and formed. In any chemical reaction, it is crucial to identify and understand the involved reactants and products.
In the given exercise, zinc (\(\mathrm{Zn}\)) reacts with silver nitrate (\(\mathrm{AgNO}_3\)) to form silver (\(\mathrm{Ag}\)) and zinc nitrate (\(\mathrm{Zn(NO}_3)_2\)). This transformation is evidenced by changes like color formation or gas evolution that indicate a reaction has occurred. Furthermore, chemical reactions can be categorized into several types, such as synthesis, decomposition, single replacement, and double replacement, to name a few. The current exercise illustrates a single replacement reaction, where zinc displaces silver fromsilver nitrate, leading to the formation of \(\mathrm{Ag}\) and \(\mathrm{Zn(NO}_3)_2\). Understanding the nature of chemical reactions helps to predict the behavior and product formation in given scenarios.

Limiting Reactant

The concept of a limiting reactant is essential in stoichiometry which governs the completion of a reaction. The limiting reactant is the substance that is entirely consumed first during a chemical reaction, thereby determining the maximum amount of product that can be formed. It limits the extent of the reaction because once it is used up, no further products can be made, regardless of the amounts of other reactants present.
In the exercise, zinc and silver nitrate are the reactants, but not both can completely react according to the balanced equation. Using stoichiometry calculations, we found that silver nitrate (\(\mathrm{AgNO}_3\)) is the limiting reactant. This is because the available moles of \(\mathrm{AgNO}_3\) are insufficient to fully react with all the available moles of \(\mathrm{Zn}\). This determines the amount of silver metal (\(\mathrm{Ag}\)) that is formed, as the reaction ceases when \(\mathrm{AgNO}_3\) is exhausted.

Molar Mass Calculations

Molar mass calculations are fundamental in stoichiometry to convert between the mass of a substance and the amount in moles. The molar mass is the mass of a given substance (molecule, compound, element) present in one mole, typically expressed in grams per mole (g/mol). To calculate it, you sum the atomic masses of each element present in the molecule or compound.
For our exercise:

• Zinc (\(\mathrm{Zn}\)) has a molar mass of 65.38 g/mol, calculated from its atomic weight.
• Silver nitrate (\(\mathrm{AgNO}_3\)) has a molar mass of 169.87 g/mol, derived from adding the atomic masses of Ag, N, and three O atoms.
• Silver (\(\mathrm{Ag}\)) has a molar mass of 107.87 g/mol.
• Zinc nitrate (\(\mathrm{Zn(NO}_3)_2\)) has a molar mass of 189.36 g/mol.

These values are used to calculate how many moles correspond to a given mass of the substance, facilitating stoichiometric calculations like finding how much of each product forms given the starting amounts of reactants.

Balanced Chemical Equation

A balanced chemical equation is crucial in stoichiometry because it obeys the law of conservation of mass. This law states that in a chemical reaction, matter cannot be created or destroyed. Therefore, a balanced equation has the same number of each type of atom on both sides of the equation.
In the exercise, the initial equation was unbalanced:\(\mathrm{Zn}(s) + \mathrm{AgNO}_3(aq) \rightarrow \mathrm{Ag}(s) + \mathrm{Zn(NO}_3)_2(aq)\)To balance it, we ensure that the same number of atoms for each element are present on both sides of the equation. This was achieved by adding the coefficient 2 in front of \(\mathrm{AgNO}_3\) and \(\mathrm{Ag}\), resulting in:\(\mathrm{Zn}(s) + 2\mathrm{AgNO}_3(aq) \rightarrow 2\mathrm{Ag}(s) + \mathrm{Zn(NO}_3)_2(aq)\)Balancing equations provides the exact proportions needed for reactions to proceed correctly, allowing accurate calculations of the substances involved. Without this step, stoichiometric calculations of product yields and limiting reactants would be ineffective.