Question

Consider the reaction of \(19.0 \mathrm{~g}\) of zinc with excess silver nitrite to produce silver metal and zinc nitrite. The reaction is stopped before all the zinc metal has reacted and \(29.0 \mathrm{~g}\) of solid metal is present. Calculate the mass of each metal in the 29.0-g mixture.

Short answer

In the 29.0 g mixture, there are \(3.319\,\text{g}\) of unreacted zinc and \(25.681\,\text{g}\) of silver.

Step-by-step solution

Write the balanced chemical equation for the reaction

For the given reaction, zinc (Zn) reacts with silver nitrite (AgNO₂) to produce silver (Ag) and zinc nitrite (Zn(NO₂)₂). So, the balanced chemical equation would be: \[Zn + 2AgNO_2 \rightarrow 2Ag + Zn(NO_2)_2\]

Find the moles of zinc given

We are given 19.0 g of zinc. To find the number of moles of zinc, we will use the molar mass of zinc, which is 65.38 g/mol. So, the moles of zinc will be: \[\frac{19.0\,\text{g}}{65.38\,\text{g/mol}} = 0.2907\,\text{mol} \]

Determine the mass of silver produced

From the balanced chemical equation in Step 1, we see that 1 mole of zinc reacts with 2 moles of silver nitrite to produce 2 moles of silver. So, the moles of silver produced will be equal to the moles of zinc. Using the moles of zinc calculated in Step 2, and the molar mass of silver (107.87 g/mol), we can find the mass of silver produced: \[0.2907\,\text{mol} \times 107.87\,\text{g/mol} = 31.36\,\text{g}\] However, the total mass of solid metal is given to be 29.0 g, which means the reaction did not go to completion and not all the zinc reacted.

Calculate the mass of unreacted zinc

If all the zinc had reacted, we would have had 31.36 g of silver. But since we only have 29.0 g of solid metal mixture, we need to find out how much of the calculated silver was actually produced and subtract that from the total mass to find the mass of unreacted zinc. Let x be the mass of the unreacted zinc, then the mass of actually produced silver is (29.0 - x) g. From the stoichiometry of the reaction, we can write: \[\frac{29.0-x\,\text{g}}{107.87\,\text{g/mol}} = \frac{0.2907\,\text{mol}}{2}\] Solving for x, we get: \[x = 19.0 - \left( \frac{0.2907\,\text{mol}}{2} \times 107.87\,\text{g/mol} \right) = 3.319\,\text{g}\]

Find the mass of each metal in the mixture

Now, we know the mass of unreacted zinc (3.319 g), and we can calculate the mass of silver in the mixture by subtracting the mass of unreacted zinc from the total mass (29.0 g): \[mass\,of\,Ag = 29.0\,\text{g} - 3.319\,\text{g} = 25.681\,\text{g}\] So, in the 29.0 g mixture, there are 3.319 g of unreacted zinc and 25.681 g of silver.

Key concepts

Stoichiometry

Stoichiometry is the study of quantitative relationships in a chemical reaction. It involves using balanced chemical equations to determine the proportions of reactants and products involved in a reaction. With stoichiometry, we can calculate how much of each substance is needed or produced in a reaction.

In stoichiometry, the coefficients in a chemical equation represent the ratios of moles of each substance. For example, the balanced equation for the reaction between zinc and silver nitrite is:\[Zn + 2AgNO_2 \rightarrow 2Ag + Zn(NO_2)_2\] This indicates that one mole of zinc reacts with two moles of silver nitrite to produce two moles of silver and one mole of zinc nitrite. By understanding these relationships, we can calculate the expected amount of products formed and the reactants needed, helping us solve problems like determining limiting reactants and reaction yields.

Balanced Chemical Equation

A balanced chemical equation is crucial for accurately representing a chemical reaction. It ensures that the same number of each type of atom appears on both sides of the equation, satisfying the law of conservation of mass. This means that atoms are neither created nor destroyed in a reaction, so the mass remains constant.

In our exercise, the balanced equation is: \[Zn + 2AgNO_2 \rightarrow 2Ag + Zn(NO_2)_2\] Here, we see that both sides of the equation have the same number of atoms for each element involved. Balancing chemical equations requires adjusting the coefficients in front of the reactant and product formulas without changing the actual chemical formulas. This step is fundamental for making accurate stoichiometric calculations in chemistry.

Molar Mass

Molar mass is the mass of one mole of a given substance, typically measured in g/mol. It links the mass of a substance to the amount in moles, allowing chemists to convert between the two. The molar mass can be found by summing the atomic masses of all the atoms in a molecule's chemical formula.

For the calculation involving zinc in our problem, the molar mass of zinc is 65.38 g/mol. This value helps us convert the given mass of zinc into moles:\[\frac{19.0\,\text{g}}{65.38\,\text{g/mol}} = 0.2907\,\text{mol}\] Similarly, the molar mass of silver, 107.87 g/mol, is used to calculate the expected mass of silver produced from the moles of zinc:\[0.2907\,\text{mol} \times 107.87\,\text{g/mol} = 31.36\,\text{g}\] Understanding molar mass is essential for connecting the macroscopic world of grams to the microscopic world of moles in chemical reactions.

Reaction Completion

Reaction completion refers to the extent to which reactants are converted into products in a chemical reaction. A reaction is deemed complete when all reactants are used up, meaning none are left unreacted. However, in some cases, not all reactants react completely, often due to limitations in the reaction conditions, like time or concentration of the reactants.

In our exercise, the reaction between zinc and silver nitrite did not go to completion. This is evident since there is an observed discrepancy between the calculated mass of silver produced (31.36 g) and the actual mass of the metal mixture (29.0 g).

• This indicates some zinc was left unreacted.
• The calculation showed 3.319 g of zinc was unreacted.
• 25.681 g of the mixture is actually silver.

By understanding the reaction completion, chemists can analyze and optimize reactions, ensuring that reactions proceed as fully as possible to maximize yield.