/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 121 Douglasite is a mineral with the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Douglasite is a mineral with the formula 2 \(\mathrm{KCl} \cdot \mathrm{FeCl}_{2}\) . 2 \(\mathrm{H}_{2} \mathrm{O}\) . Calculate the mass percent of douglasite in a 455.0 \(\mathrm{-mg}\) sample if it took 37.20 \(\mathrm{mL}\) of a \(0.1000-M \mathrm{AgNO}_{3}\) solution to precipitate all the \(\mathrm{Cl}^{-}\) as AgCl. Assume the douglasite is the only source of chloride ion.

Short Answer

Expert verified
The mass percent of douglasite in the 455.0 mg sample is approximately 112.64%.

Step by step solution

01

Calculate the moles of AgNO3

To calculate the moles of AgNO3, we will use the given volume and molarity of the solution: \(Moles\, of\, AgNO_3 = Volume\, of\, solution × Molarity\) Moles of AgNO3 = \(37.20 \, mL × 0.1000 \, M\) Note: Convert the volume from mL to L Moles of AgNO3 = \(0.03720 \, L × 0.1000 \, M = 3.72 × 10^{-3} \, moles\)
02

Calculate the moles of Cl- ions in the sample

Since AgCl is formed from the reaction of AgNO3 with Cl- ions, the number of moles of Cl- ions will be equal to the number of moles of AgNO3. Moles of Cl- ions = Moles of AgNO3 = \(3.72 × 10^{-3} \, moles\)
03

Calculate the moles of douglasite

Looking at the formula for douglasite (2 KCl·FeCl2), there are three moles of Cl- ions in one mole of douglasite. We will use this molar ratio to calculate the moles of douglasite. Moles of douglasite = \(\frac{moles\, of\, Cl-}{moles\, of\, Cl- per\, mole\, of\, douglasite}\) Moles of douglasite = \(\frac{3.72 × 10^{-3} \, moles}{3}\) Moles of douglasite = \(1.24 × 10^{-3} \, moles \)
04

Calculate the mass of douglasite present in the sample

To calculate the mass of douglasite, we need to multiply the moles of douglasite by its molar mass: Molar mass of douglasite ≈ (2 × M(KCl) + M(FeCl2) + 2 × M(H2O)) Molar mass of douglasite ≈ 2 × (39.10 + 35.45) + (55.85 + 2 × 35.45) + 2 × (2 × 1.01 + 16.00) g/mol ≈ 413.7 g/mol Now, calculating the mass of douglasite in the sample: Mass of douglasite = Moles of douglasite × Molar mass of douglasite Mass of douglasite = \(1.24 × 10^{-3} \, moles × 413.7 \, \frac{g}{mol} ≈ 0.5125 \, g\)
05

Calculate the mass percent of douglasite in the sample

Finally, we will calculate the mass percent of douglasite in the sample: Mass percent of douglasite = \(\frac{Mass\, of\, douglasite}{Mass\, of\, sample} × 100\) Mass percent of douglasite = \(\frac{0.5125 \, g}{0.4550 \, g} × 100 ≈ 112.64\%\) In conclusion, the mass percent of douglasite in the 455.0 mg sample is approximately 112.64%.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Moles of AgNO3
To find the moles of a substance, you multiply its volume by its molarity. This is especially useful in reactions involving a solution, like with \(\mathrm{AgNO_3}\)in this case. When a problem provides molarity and a specific volume of a solution, you start by converting the volume from milliliters to liters. This step is crucial since molarity is defined in terms of liters:
  • Conversion: \(37.20 \, \mathrm{mL} = 0.03720 \, \mathrm{L}\)
Next, use the formula\[ Moles\, of\, \mathrm{AgNO_3} = \mathrm{Volume\, (L) } \times \mathrm{Molarity\, (M)} \ = 0.03720\, \mathrm{L} \times 0.1000\, \mathrm{M} = 3.72 \times 10^{-3} \mathrm{moles}\]You've now got the moles of \(\mathrm{AgNO_3}\). Understanding this means having insight into the amount of substance present, which is vital for further calculations.
Precipitation of AgCl
When \(\mathrm{AgNO_3}\) is mixed with chloride ions in a solution, a chemical reaction occurs, forming insoluble silver chloride (AgCl) as a precipitate. This means that all available chloride ions will react with \(\mathrm{AgNO_3}\) until one of the reactants is used up. The stoichiometry of the reaction follows a 1:1 ratio:
  • Each mole of \(\mathrm{Cl^-}\) requires one mole of \(\mathrm{AgNO_3}\) to completely precipitate as \(\mathrm{AgCl}\).
Thus, the number of moles of chloride ions will be equal to the moles of \(\mathrm{AgNO_3}\) used, which we've already calculated to be\[3.72 \times 10^{-3} \mathrm{\, moles}\]This allows us to conclude the presence of \(\mathrm{Cl^-}\) ions in the original sample. This step is critical because it links the silver solution used to the chloride ions in the sample.
Molar Mass of Douglasite
To proceed with the mass calculations, the molar mass of the compound, such as douglasite, becomes essential. Douglasite has the formula \(2\ \mathrm{KCl} \cdot \mathrm{FeCl}_2 \cdot 2\ H_2O\). Summing up the atomic masses gives the molar mass. Let's break it down:
  • For \(\mathrm{KCl}:\)
    • Potassium (K): \(39.10 \, \mathrm{g/mol}\)
    • Chlorine (Cl): \(35.45 \, \mathrm{g/mol}\)
    \(2 \times (39.10 + 35.45)\)
  • For \(\mathrm{FeCl}_2:\)
    • Iron (Fe): \(55.85 \, \mathrm{g/mol}\)
    • Chlorine: \(2 \times 35.45 \, \mathrm{g/mol}\)
  • Water, \(\mathrm{H_2O}:\) \(2 \times (2 \times 1.01 + 16.00)\)
When added together, the molar mass of douglasite approximates to \[\mathrm{413.7 \, g/mol}\]This large number signifies the weight of one mole of douglasite, which is vital when calculating the mass of douglasite from its molar quantity.
Chloride Ions in Minerals
Chloride ions are an essential component in many mineral forms. In this problem, douglasite is our source of \(\mathrm{Cl^-}\) ions. When you analyze a sample for its chloride content, underlying reactions involve its conversion into something measurable, such as \(\mathrm{AgCl}\) precipitate.In minerals like douglasite, the formula indicates the chloride content. Two moles of \(\mathrm{KCl}\) and one of \(\mathrm{FeCl_2}\) suggest three chloride ions per douglasite molecule. This formula tells us there is a fixed proportion of chloride within every mole of the mineral.This knowledge allows chemists to compute the potential release of chloride ions when the mineral is dissolved or altered chemically. Comprehending the amount of chloride ions helps in determining the mineral's composition and its potential usage in various chemical reactions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You are given a 1.50-g mixture of sodium nitrate and sodium chloride. You dissolve this mixture into 100 mL of water and then add an excess of 0.500 M silver nitrate solution. You produce a white solid, which you then collect, dry, and measure. The white solid has a mass of 0.641 g. a. If you had an extremely magnified view of the solution (to the atomic- molecular level), list the species you would see (include charges, if any). b. Write the balanced net ionic equation for the reaction that produces the solid. Include phases and charges. c. Calculate the mass percent of sodium chloride in the original unknown mixture.

Hydrochloric acid \((75.0 \mathrm{mL} \text { of } 0.250 \mathrm{M})\) is added to 225.0 \(\mathrm{mL}\) of 0.0550 \(\mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution. What is the concentration of the excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in this solution?

What volume of 0.0521\(M \mathrm{Ba}(\mathrm{OH})_{2}\) is required to neutralize exactly 14.20 \(\mathrm{mL}\) of 0.141 \(\mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4} ?\) Phosphoric acid contains three acidic hydrogens.

The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) level can be determined by tirrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of \(\mathrm{Cr}^{3+}(a q)\) and carbon dioxide. The reaction can be monitored because the dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right)\) is orange in solution, and the \(\mathrm{Cr}^{3+}\) ion is green. The unbalanced redox equation is $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{CO}_{2}(g)$$ If 31.05 \(\mathrm{mL}\) of 0.0600\(M\) potassium dichromate solution is required to titrate 30.0 \(\mathrm{g}\) of blood plasma, determine the mass percent of alcohol in the blood.

A 25.00-mL sample of hydrochloric acid solution requires 24.16 mL of 0.106 M sodium hydroxide for complete neutralization. What is the concentration of the original hydrochloric acid solution?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.