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A 450.0 -mL sample of a \(0.257 M\) solution of silver nitrate is mixed with \(400.0 \mathrm{~mL}\) of 0.200 \(M\) calcium chloride. What is the concentration of \(\mathrm{Cl}^{-}\) in solution after the reaction is complete?

Short Answer

Expert verified
The concentration of Cl鈦 ions in the final solution after the reaction is complete is \(0.05218 \mathrm{M}\).

Step by step solution

01

Write the balanced chemical equation

For the reaction between silver nitrate (AgNO鈧) and calcium chloride (CaCl鈧), the balanced chemical equation is: \(2 AgNO_3 + CaCl_2 \rightarrow 2 AgCl \downarrow + Ca(NO_3)_2\) In this reaction, 2 moles of silver nitrate react with 1 mole of calcium chloride to form 2 moles of silver chloride solid precipitate and 1 mole of calcium nitrate.
02

Calculate the moles of each reactant

Next, we will need to determine the moles of each reactant present in the given volumes and concentrations: Moles of AgNO鈧 = (Volume of AgNO鈧) 脳 (Concentration of AgNO鈧) Moles of AgNO鈧 = (450.0 mL) 脳 (0.257 mol/L) Convert the volume to liters before calculating the moles: Moles of AgNO鈧 = (0.450 L) 脳 (0.257 mol/L) = 0.11565 mol AgNO鈧 Moles of CaCl鈧 = (Volume of CaCl鈧) 脳 (Concentration of CaCl鈧) Moles of CaCl鈧 = (400.0 mL) 脳 (0.200 mol/L) Convert the volume to liters before calculating the moles: Moles of CaCl鈧 = (0.400 L) 脳 (0.200 mol/L) = 0.08000 mol CaCl鈧
03

Determine the limiting reactant

Now, we will need to determine the limiting reactant by comparing the moles of each reactant present with the stoichiometry of the balanced chemical equation: Ratio of AgNO鈧 / CaCl鈧 in the balanced equation = 2 / 1 Divide the moles of each reactant by the coefficients in the balanced equation: 0.11565 mol AgNO鈧 / 2 = 0.057825 0.08000 mol CaCl鈧 / 1 = 0.080 As 0.057825 < 0.080, AgNO鈧 is the limiting reactant in this reaction.
04

Calculate the moles of Cl鈦 ions remaining after the reaction

Since AgNO鈧 is the limiting reactant, all 0.11565 mol of AgNO鈧 will react with the available CaCl鈧. To determine the remaining moles of Cl鈦 ions after the reaction, we can use the stoichiometry of the balanced chemical equation: 0.11565 mol AgNO鈧 脳 (1 mol CaCl鈧 / 2 mol AgNO鈧) = 0.057825 mol CaCl鈧 reacted Since we initially had 0.08000 mol of CaCl鈧, we can calculate the remaining moles: 0.08000 mol CaCl鈧 - 0.057825 mol reacted = 0.022175 mol Cl鈦 ions remaining after the reaction (since each CaCl鈧 has 2 moles of Cl鈦 ions, multiply by 2) Remaining moles of Cl鈦 ions = 0.022175 mol 脳 2 = 0.04435 mol Cl鈦 ions
05

Calculate the final concentration of Cl鈦 ions

Now we can calculate the concentration of Cl鈦 ions in the solution. First, we need to find the total volume of the solution after mixing the two solutions: Total volume = Volume of AgNO鈧 + Volume of CaCl鈧 = 0.450 L + 0.400 L = 0.850 L Now, divide the remaining moles of Cl鈦 ions by the total volume of the solution: Concentration of Cl鈦 ions = 0.04435 mol Cl鈦 ions / 0.850 L = 0.05218 M The concentration of Cl鈦 ions in the final solution is 0.05218 M.

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Key Concepts

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

Stoichiometry
Stoichiometry is a section of chemistry that involves using relationships from balanced chemical equations to calculate quantities of reactants or products. It is analogous to using a recipe in cooking where you figure out how much of each ingredient you need to serve a certain number of people.

In our example, stoichiometry is used to determine the moles of reactants (silver nitrate and calcium chloride) using their concentrations and volumes. This conversion from volume to moles is central to the concept because reactions occur at the molecular level, and stoichiometry allows us to work at this scale. By understanding how substances react in fixed ratios, as dictated by the coefficients in a balanced chemical equation, we can predict the amounts of products formed and reactants used up.

Critical to this is the mole, which is the unit of amount in chemistry. Stoichiometry enables us to convert between moles, mass, volume, and particles, making it a powerful tool for scientists to predict the outcomes of chemical reactions. In practical terms, knowing stoichiometry can help us calculate how much of a substance we need to make a particular product or how much product we can expect to be formed from a given amount of a reactant.
Limiting Reactant
The limiting reactant in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The quantity of the limiting reactant determines the maximum amount of product that can be formed; it 'limits' the reaction. Determining the limiting reactant is a key step in stoichiometry.

In our exercise, the comparison between the amount of moles of silver nitrate and calcium chloride, adjusted for their stoichiometric coefficients in the balanced equation, shows that silver nitrate is the limiting reactant. In the kitchen, if you're baking cookies and run out of sugar, sugar is your limiting 'reactant,' and no matter how much of the other ingredients you have, you can't make more cookies than the sugar allows. Similarly, once all the silver nitrate has reacted, the reaction stops, regardless of how much calcium chloride remains.

Understanding the concept of the limiting reactant prevents wasting other reactants and resources in practical applications, such as in industrial chemical syntheses, pharmaceuticals, consumer goods manufacturing, and even in environmental control when treating waste.
Molarity
Molarity, denoted as 'M,' is a unit of concentration used in chemistry to express the number of moles of a solute per liter of solution. It allows chemists to communicate how 'strong' or 'concentrated' a solution is. The formula for molarity is \( M = \frac{mol}{L} \).

The calculation of molarity is an essential skill in chemistry because it directly relates the number of particles in a given volume, providing a way to understand the concentration of substances in solution. In our textbook exercise, the students start with molar concentrations of two solutions and must calculate the molarity of chloride ions after the reaction is complete. This type of calculation is not only academically relevant but also applicable in real-life situations, such as adjusting the concentration of a medicinal prescription or determining the amount of a reagent needed for a biological assay.

When two solutions of silver nitrate and calcium chloride are mixed, the volume changes, which affects the final molarity of the products in the solution. Hence, converting moles back to molarity is crucial for finding out the concentration of chloride ions remaining post-reaction. This ability to play with and understand concentrations defines a great portion of work in fields like biochemistry, pharmacology, and material science.

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Most popular questions from this chapter

An alcoholic iodine solution ("tincture" of iodine) is prepared by dissolving 5.15 g of iodine crystals in enough alcohol to make a volume of \(225 \mathrm{~mL}\). Calculate the molarity of iodine in the solution.

Many metal ions are precipitated from solution by the sulfide ion. As an example, consider treating a solution of copper(II) sulfate with sodium sulfide solution: $$ \mathrm{CuSO}_{4}(a q)+\mathrm{Na}_{2} \mathrm{~S}(a q) \rightarrow \mathrm{CuS}(s)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) $$ What volume of \(0.105 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}\) solution would be required to precipitate all of the copper(II) ion from \(27.5 \mathrm{~mL}\) of \(0.121 \mathrm{M} \mathrm{CuSO}_{4}\) solution?

Consider the reaction between \(1.0 \mathrm{~L}\) of \(3.0 \mathrm{M} \mathrm{AgNO}_{3}(a q)\) and \(1.0 \mathrm{~L}\) of \(1.0 \mathrm{M} \mathrm{CuCl}_{2}(a q)\) according to the equation: $$ 2 \mathrm{AgNO}_{3}(a q)+\mathrm{CuCl}_{2}(a q) \rightarrow 2 \mathrm{AgCl}(s)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q) $$ Which of the following will increase the amount of AgCl (s) produced? a. Adding another \(1.0 \mathrm{~L}\) of \(3.0 \mathrm{M} \mathrm{AgNO}_{3}(a q)\) b. Adding another \(1.0 \mathrm{~L}\) of \(1.0 \mathrm{M} \mathrm{CuCl}_{2}(a q)\) c. Adding \(1.0 \mathrm{~L}\) of water to the original \(\mathrm{AgNO}_{3}(a q)\) solution d. Allowing all of the water to evaporate from the resulting solution e. At least two of the above will increase the amount of \(\mathrm{AgCl}(s)\) produced.

What volume of \(0.250 \mathrm{M} \mathrm{HCl}\) is required to neutralize each of the following solutions? a. \(25.0 \mathrm{~mL}\) of \(0.103 \mathrm{M}\) sodium hydroxide, \(\mathrm{NaOH}\) b. \(50.0 \mathrm{~mL}\) of \(0.00501 M\) calcium hydroxide, \(\mathrm{Ca}(\mathrm{OH})_{2}\) c. \(20.0 \mathrm{~mL}\) of \(0.226 \mathrm{M}\) ammonia, \(\mathrm{NH}_{3}\) d. \(15.0 \mathrm{~mL}\) of \(0.0991 \mathrm{M}\) potassium hydroxide, \(\mathrm{KOH}\)

Calcium oxalate, \(\mathrm{CaC}_{2} \mathrm{O}_{4},\) is very insoluble in water. What mass of sodium oxalate, \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) is required to precipitate the calcium ion from \(37.5 \mathrm{~mL}\) of \(0.104 \mathrm{M} \mathrm{CaCl}_{2}\) solution?

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