A solution contains Cr 3+ ions and Mg 2+ ions. The addition of 1.00 L of 1.51 M NaF solution causes the complete precipitation of these ions as CrF 3 (s) and MgF 2 (s). The total mass of the precipitate is 49.6 g. Find the mass of Cr 3+ in the original solution.

Hi everyone here we have a question asking us to consider a solution containing a mixture of silver and lead ions. The ions were precipitated out from the solution as silver chloride solid and lead chloride solid by adding 1.0 liters of 0.250 moller potassium chloride. The precipitate was collected and had a mass of 35.4 g. And our goal is to find the mass of lead in the initial solution. So we're going to use more clarity equals moles over leaders of solution. So our balanced equation that we're going to use is potassium chloride dissociates into potassium ions plus chloride ions. So our moles of chloride equals 1. liters times 0.250 moles of potassium chloride over leaders, times one mole of chlorine over one mole a potassium chloride. And our leaders are going to cancel out our moles of potassium chloride are going to cancel out. And that's going to give us 0. moles of chloride. So we're going to let x equal moles of silver chloride. And why equal moles of lead chloride? Which means that our moles of chloride from potassium or from silver chloride is going to equal X and are moles of chloride from lead chloride is going to equal to Y. Because we have to chlorine and lead chloride. So X plus two Y. Is therefore going to equal our total moles of chloride which is 0.250 moles Of chloride. And now we need to isolate. Why? So we're going to -1 from both sides. So two y. is going to equal 0. moles of chloride minus X. And then we're gonna divide both sides by two. So why equals 0.250 minus X, divided by two. Now we're going to calculate our molar mass of silver chloride. So that's going to equal 107.87 g per mole. And that is for silver and that's just found on the periodic table plus 35.45 g per mole. And that's our chlorine. And again, it's just found on the periodic table and that equals 143.32 g per mole. So our mass of silver chloride equals x moles of silver chloride Times 143.32 g Permal, So that equals 143. x. And now we're going to do the same for our lead chloride. So our molar mass of lead chloride Is going to equal 207.2 g per mole. And that is for our lead Plus two times 35.45 g per mole. And that is for our chlorine. And that equals 278.1 g per mole. So our mass of lead chloride is going to equal y moles of lead chloride Because we don't know how many moles yet times are Molar mass, which is 278.1 g per mole that we just calculated. So that's going to equal what we calculated before. So 0. -1 divided by two Times 278.1. So our total mass is going to equal our mass of silver chloride plus our mass of lead chloride. So that means 35. g Is going to equal 143.32 x plus 34. -139. x. So 35 .4g equals 143 . x plus 34. -139.05 x. So 35.4 equals 4.27 x plus 34 .7625. So 35. -34. equals 4.27 x. So 0.6375 equals 4.2754 point X. And then we divide by the 4.27. So x equals 0.1493. And now we're going to plug that back in to our y equation. So why equals 0. -0.1493, divided by two Equals 0.05035 moles of lead chloride. And now finally, we need to calculate the mass of lead. So our molar mass of lead found on the periodic table is 207.2 g per mole. So our mass of lead equals 0. mold of lead chloride times one mole of lead over one mole of lead chloride times 207. g Permal. So our moles of lead chloride are canceling out and our moles of lead are canceling out, and that gives us 10.4 g. And that is our final answer. Thank you for watching. Bye.

Ch.4 - Chemical Quantities & Aqueous Reactions

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Tro 4th Edition

Ch.4 - Chemical Quantities & Aqueous Reactions

Problem 116

Chapter 4, Problem 116

A solution contains Cr³⁺ ions and Mg²⁺ ions. The addition of 1.00 L of 1.51 M NaF solution causes the complete precipitation of these ions as CrF₃(s) and MgF₂(s). The total mass of the precipitate is 49.6 g. Find the mass of Cr³⁺ in the original solution.

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Determine the moles of NaF added using the concentration and volume: \( \text{moles of NaF} = 1.51 \text{ M} \times 1.00 \text{ L} \).

Write the balanced chemical equations for the precipitation reactions: \( \text{Cr}^{3+} + 3\text{F}^- \rightarrow \text{CrF}_3(s) \) and \( \text{Mg}^{2+} + 2\text{F}^- \rightarrow \text{MgF}_2(s) \).

Calculate the moles of \( \text{CrF}_3 \) and \( \text{MgF}_2 \) that can be formed using the stoichiometry of the reactions and the moles of NaF available.

Use the molar masses of \( \text{CrF}_3 \) and \( \text{MgF}_2 \) to express the total mass of the precipitate in terms of the moles of \( \text{CrF}_3 \) and \( \text{MgF}_2 \).

Set up an equation for the total mass of the precipitate (49.6 g) and solve for the moles of \( \text{CrF}_3 \), then calculate the mass of \( \text{Cr}^{3+} \) in the original solution.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It involves using balanced chemical equations to determine the relationships between the amounts of substances consumed and produced. In this question, stoichiometry is essential for relating the moles of Cr3+ and Mg2+ ions to the moles of their respective precipitates, CrF3 and MgF2.

Precipitation Reactions

Precipitation reactions occur when two soluble salts react in solution to form an insoluble solid, known as a precipitate. In this case, the addition of NaF leads to the formation of CrF3 and MgF2 precipitates. Understanding the solubility rules and the conditions under which these ions precipitate is crucial for determining the mass of the original ions in the solution.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is used to convert between the mass of a substance and the number of moles. In this problem, calculating the mass of Cr3+ requires knowing its molar mass to relate the total mass of the precipitate to the moles of Cr3+ that contributed to the formation of CrF3.