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Chapter 6: Problem 127

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

Expert verified
In the 29.0 g mixture, there are 9.0 g of zinc remaining and 20.0 g of silver produced.

Step by step solution

01

Write the balanced chemical equation for the reaction

The reaction between zinc metal and silver nitrite can be represented as: \[Zn + 2AgNO_2 \rightarrow 2Ag + Zn(NO_2)_2\]
02

Calculate the moles of zinc initially present

To calculate the moles of zinc initially present in the reaction, use the given mass and the molar mass of zinc (65.38 g/mol): Moles of zinc = \(\frac{mass}{molar\ mass}\) Moles of zinc = \(\frac{19.0 \ g}{65.38 \ g/mol}\) = 0.2906 mol
03

Determine the maximum amount of silver that can be produced

According to the balanced chemical equation, 1 mole of zinc can produce 2 moles of silver. Therefore, we can calculate the maximum moles of silver that can be produced: Maximum moles of silver = moles of zinc 脳 \(2\) Maximum moles of silver = 0.2906 mol 脳 \(2\) = 0.5812 mol Now, we can find the mass of the maximum amount of silver: Mass of maximum silver = moles 脳 molar mass Mass of maximum silver = 0.5812 mol 脳 \(107.87 \ g/mol\) = 62.7 g
04

Determine the mass of unreacted zinc

Since the reaction stopped before all the zinc was consumed, some zinc remains unreacted. We know that the total mass of solid metal at the end of the reaction is 29.0 g, so we can find the mass of unreacted zinc: Mass of unreacted zinc = Total mass of solid metal - Mass of maximum silver Mass of unreacted zinc = \(29.0 \ g - 62.7 \ g\) = -33.7 g The negative value indicates that there is less silver produced than the maximum possible amount since not all the zinc was consumed, and we should recalculate for this case.
05

Determine the actual mass of silver produced and zinc remaining

Assume 'x' grams of zinc reacted. Then, according to the stoichiometry, 2 times the mass of silver is produced: Mass of silver = \(2x\) Mass of remaining zinc = \(19.0 \ g - x\) The total mass of solid metal in the mixture is given as 29.0 g: Total mass of solid metal = Mass of silver + Mass of remaining zinc 29.0 g = \(2x + 19.0 \ g - x\) Now, solving for 'x': x = 10.0 g So, 10.0 g of zinc reacted, and the mass of the remaining zinc is: Mass of remaining zinc = \(19.0 \ g - 10.0 \ g\) = 9.0 g The mass of silver produced is: Mass of silver = \(2 脳 10.0 g\) = 20.0 g
06

Conclusion

In the 29.0 g mixture, there are 9.0 g of zinc remaining and 20.0 g of silver produced.

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Key Concepts

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

Balanced Chemical Equation
Understanding balanced chemical equations is crucial in the study of chemical reactions. In our exercise, the chemical reaction shows zinc reacting with silver nitrite.
The balanced chemical equation is: \[Zn + 2AgNO_2 \rightarrow 2Ag + Zn(NO_2)_2\]
This equation is balanced because the number of atoms for each element is the same on both sides of the equation. Here's why a balanced equation is important:
  • It ensures that the law of conservation of mass is upheld, meaning that mass is neither created nor destroyed during a chemical reaction.
  • The coefficients in front of each compound indicate the molar ratio of the reactants and products, which is essential for calculating other properties like mass and moles.
A balanced equation helps us predict the amounts of products formed from given reactants, understanding the stoichiometric relationships, which is the backbone of solving problems related to chemical reactions.
Mole Calculation
Mole calculations help convert between the mass of a substance and the number of particles or entities it contains. In this exercise, we calculated the moles of zinc:
Using the equation: \[\text{Moles of zinc} = \frac{19.0 \ g}{65.38 \ g/mol} = 0.2906 \text{ moles}\]
Mole calculation is important for several reasons:
  • It allows us to use the mole ratios from a balanced chemical equation to determine how much of a product can be formed from a reactant.
  • By knowing moles, we can easily convert to other units, such as molecules, using Avogadro's number.
  • Mole calculations simplify complex chemical reactions by focusing on the amount of substance.
It's the universal way chemists connect the measurable amounts of substances to the atomic and molecular level, making it a fundamental part of chemical problem-solving.
Mass Calculation
Mass calculation involves determining the mass of reactants or products in a chemical reaction. In our problem, we found both the reacted zinc and produced silver by understanding their masses in the equation:
  • The mass of reacted zinc: \(10.0 \ g\)
  • The mass of silver produced: \(20.0 \ g\)
  • The mass of remaining zinc: \(9.0 \ g\)
Mass calculation steps:
  • Calculate the maximum theoretical amount of a product using mole ratios from a balanced equation.
  • Apply known masses and ratios to find unknowns, as seen when recalculating the actual mass of products formed when the reaction doesn't proceed to completion.
By combining stoichiometry with mass calculations, we analyze real lab scenarios, ensuring accurate measurements and predictions.

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Most popular questions from this chapter

Which of the following solutions of strong electrolytes contains the largest number of moles of chloride ions: \(100.0 \mathrm{mL}\) of \(0.30 \mathrm{M} \mathrm{AlCl}_{3}, 50.0 \mathrm{mL}\) of \(0.60 \mathrm{M} \mathrm{MgCl}_{2},\) or \(200.0 \mathrm{mL}\) of \(0.40 M\) NaCl?

The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, I ppm means 1 part of solute for every \(10^{6}\) parts of solution. Mathematically, by mass: $$ \mathrm{ppm}=\frac{\mu \mathrm{g} \text { solute }}{\mathrm{g} \text { solution }}=\frac{\mathrm{mg} \text { solute }}{\mathrm{kg} \text { solution }} $$ In the case of very dilute aqueous solutions, a concentration of \(1.0 \mathrm{ppm}\) is equal to \(1.0 \mu \mathrm{g}\) of solute per \(1.0 \mathrm{mL},\) which equals 1.0 g solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. \(5.0 \mathrm{ppb} \mathrm{Hg}\) in \(\mathrm{H}_{2} \mathrm{O}\) b. \(1.0 \mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\) c. \(10.0 \mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\) d. \(0.10 \mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}\right)\) in \(\mathrm{H}_{2} \mathrm{O}\)

Many oxidation-reduction reactions can be balanced by inspection. Try to balance the following reactions by inspection. In each reaction, identify the substance reduced and the substance oxidized. a. \(\mathrm{Al}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{CH}_{4}(g)+\mathrm{S}(s) \rightarrow \mathrm{CS}_{2}(l)+\mathrm{H}_{2} \mathrm{S}(g)\) c. \(C_{3} H_{8}(g)+O_{2}(g) \rightarrow C O_{2}(g)+H_{2} O(l)\) d. \(\mathrm{Cu}(s)+\mathrm{Ag}^{+}(a q) \rightarrow \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)\)

A 0.500 -L sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution was analyzed by taking a 100.0-mL aliquot and adding \(50.0 \mathrm{mL}\) of \(0.213 \mathrm{M}\) NaOH. After the reaction occurred, an excess of \(\mathrm{OH}^{-}\) ions remained in the solution. The excess base required \(13.21 \mathrm{mL}\) of \(0.103 M\) HCI for neutralization. Calculate the molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Sulfuric acid has two acidic hydrogens.

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\operatorname{FeSO}_{4}(a q)+\operatorname{KCl}(a q)\) b. \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q)\) c. \(\mathrm{CaCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) d. \(\mathrm{K}_{2} \mathrm{S}(a q)+\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)\)
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