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Douglasite is a mineral with the formula \(2 \mathrm{KCl} \cdot \mathrm{FeCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\). Calculate the mass percent of douglasite in a \(455.0-\mathrm{mg}\) sample if it took \(37.20 \mathrm{~mL}\) of a \(0.1000 \mathrm{M} \mathrm{AgNO}_{3}\) solution to precipitate all the \(\mathrm{Cl}^{-}\) as \(\mathrm{AgCl}\). 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 48.44%, calculated by first determining the moles of Cl鈦 ions in the solution, followed by the moles of Douglasite, and then using the molar mass to determine the mass of douglasite, which finally leads to the mass percent calculation.

Step by step solution

01

Calculate moles of Cl鈦 in AgNO鈧 solution

Given the volume (37.20 mL) and concentration (0.1000 M) of AgNO鈧 solution, first, we'll calculate the moles of Cl鈦 ions in the solution. Keep in mind that 1 mole of AgNO鈧 solution reacts with 1 mole of Cl鈦 ions. Given the moles of AgNO鈧: \( moles = Molarity \times Volume \) where, Volume = 37.20 mL = 0.03720 L Molarity = 0.1000 M \( moles_{Cl^-} = 0.1000\,\,M \times 0.03720\,\,L = 0.003720\,\, mol\) So, there are 0.003720 moles of Cl鈦 ions in the AgNO鈧 solution.
02

Calculate moles of Douglasite

To calculate the moles of douglasite, we should look at its formula: \(2\,\,KCl\,\,鈰匼,\,FeCl_2\,\,鈰匼,\,2\,\,H_2O\), which has 2 moles of KCl and 1 mole of FeCl鈧, so we have a total of 2 + 2 = 4 moles of Cl鈦 ions per mole of douglasite. \(moles_{douglasite} = \frac{moles_{Cl^-}}{4}\) \(moles_{douglasite}= \frac{0.003720\,\, mol}{4} = 0.000930\,\,mol\) So, there are 0.000930 moles of douglasite.
03

Calculate the mass of Douglasite

Now, let's calculate the mass of douglasite using the molar mass and the moles of douglasite: Mass = moles 脳 molar mass The molar mass of douglasite is calculated using the atomic mass of each element multiplied by its stoichiometric coefficient in the formula: Molar mass of douglasite = 2(Molar mass of KCl) + Molar mass of FeCl鈧 + 2(Molar mass of H鈧侽) Molar mass of douglasite = 2(1 脳 39.10 + 35.45) + (55.85 + 2 脳 35.45) + 2(2 脳 1.01 +15.99) = 74.55 + 126.75 + 36.03 = 237.33 g/mol Mass of douglasite = moles 脳 molar mass Mass of douglasite = 0.000930 mol 脳 237.33 g/mol = 0.2204 g
04

Calculate mass percent of Douglasite

Now, let's calculate the mass percent of douglasite using the mass of douglasite present in the 455.0 mg sample: Mass percent of douglasite = \(\frac{mass_{douglasite}}{mass_{sample}}\) 脳 100 Since the mass of sample is given in mg, let's convert the mass of douglasite to mg as well: Mass of douglasite = 0.2204 g = 220.4 mg Mass percent of douglasite = \(\frac{220.4\,\,mg}{455.0\,\,mg}\) 脳 100 = 48.44% Therefore, the mass percent of douglasite in the 455.0 mg sample is 48.44%.

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Key Concepts

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

Stoichiometry
Stoichiometry is a branch of chemistry that deals with the quantitative relationships between the amounts of reactants and products in a chemical reaction. It provides the calculations necessary to ensure that a reaction takes place under the specified conditions. In essence, it enables us to predict the amounts of substances consumed and produced in any given chemical reaction.

When calculating stoichiometry, it's crucial to understand the concept of a mole, which is a standard unit of measurement in chemistry that represents 6.02 x 10虏鲁 particles of a substance. The balanced chemical equation provides the mole ratio between reactants and products, which serves as the basis for stoichiometric calculations. The coefficients in a chemical equation determine the ratio of moles of each substance involved, which is essential when translating chemical formulas into actual measurable quantities.

To improve the student's understanding of stoichiometry, one can focus on balancing chemical equations and converting between moles, mass, and the number of particles. This helps in mastering the fundamentals that are crucial for stoichiometry.
Molar Mass
The molar mass is the weight of one mole of a substance, typically expressed in grams per mole (g/mol). It is calculated by adding up the atomic masses of each element in the compound, as found on the periodic table, and multiplying them by their respective numbers in the compound's formula.

Understanding how to calculate the molar mass is fundamental for stoichiometry because it provides a link between the mass of a substance and the amount in moles. This is particularly useful when we need to convert grams to moles or vice versa, to measure how much of a compound is involved in a reaction.

To clarify the calculation of molar mass, we can dissect the formula of a compound into its elemental components and multiply the atomic mass of each element by the number of atoms of that element in the formula. Through practice problems, students can become proficient in calculating molar masses of various compounds, a critical step when working with mass-to-mole or mole-to-mass conversions.
Precipitation Reaction
A precipitation reaction is a type of chemical reaction where two solutions are mixed, and an insoluble solid, called a precipitate, is formed. This solid drops out of the solution and can be observed as a cloudiness or settled material at the bottom of the reaction container. Chemical equations involving precipitation reactions usually involve ionic compounds.

During a precipitation reaction, the anions and cations of the reactants exchange partners, resulting in the formation of at least one new substance that is insoluble in water. The precipitate's formula can be determined by considering the solubility rules for ionic compounds.

To reinforce the knowledge of precipitation reactions, students can be encouraged to perform lab experiments where they mix various ionic solutions and observe the formation of a precipitate. This hands-on approach not only deepens their understanding but also allows them to see the practical applications of concepts like solubility rules and ionic equations.

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Most popular questions from this chapter

A stream flows at a rate of \(5.00 \times 10^{4}\) liters per second \((\mathrm{L} / \mathrm{s})\) upstream of a manufacturing plant. The plant discharges \(3.50 \times\) \(10^{3} \mathrm{~L} / \mathrm{s}\) of water that contains \(65.0 \mathrm{ppm} \mathrm{HCl}\) into the stream. (See Exercise 113 for definitions.) a. Calculate the stream's total flow rate downstream from this plant. b. Calculate the concentration of \(\mathrm{HCl}\) in ppm downstream from this plant. c. Further downstream, another manufacturing plant diverts \(1.80 \times 10^{4} \mathrm{~L} / \mathrm{s}\) of water from the stream for its own use. This plant must first neutralize the acid and does so by adding lime: $$ \mathrm{CaO}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ What mass of \(\mathrm{CaO}\) is consumed in an \(8.00-\mathrm{h}\) work day by this plant? d. The original stream water contained \(10.2 \mathrm{ppm} \mathrm{Ca}^{2+}\). Although no calcium was in the waste water from the first plant, the waste water of the second plant contains \(\mathrm{Ca}^{2+}\) from the neutralization process. If \(90.0 \%\) of the water used by the second plant is returned to the stream, calculate the concentration of \(\mathrm{Ca}^{2+}\) in ppm downstream of the second plant.

Tris(pentafluorophenyl)borane, commonly known by its acronym BARF, is frequently used to initiate polymerization of ethylene or propylene in the presence of a catalytic transition metal compound. It is composed solely of \(\mathrm{C}, \mathrm{F}\), and \(\mathrm{B}\); it is \(42.23 \% \mathrm{C}\) by mass and \(55.66 \% \mathrm{~F}\) by mass. a. What is the empirical formula of BARF? b. A \(2.251-g\) sample of BARF dissolved in \(347.0 \mathrm{~mL}\) of solution produces a \(0.01267 M\) solution. What is the molecular formula of BARF?

What volume of \(0.0521 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) 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.

Many plants are poisonous because their stems and leaves contain oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) or sodium oxalate \(\left(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\); when ingested, these substances cause swelling of the respiratory tract and suffocation. A standard analysis for determining the amount of oxalate ion \(\left(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right)\) in a sample is to precipitate this species as calcium oxalate, which is insoluble in water. Write the net ionic equation for the reaction between sodium oxalate and calcium chloride \(\left(\mathrm{CaCl}_{2}\right)\) in aqueous solution.

The unknown acid \(\mathrm{H}_{2} \mathrm{X}\) can be neutralized completely by \(\mathrm{OH}^{-}\) according to the following (unbalanced) equation: $$ \mathrm{H}_{2} \mathrm{X}(a q)+\mathrm{OH}^{-} \longrightarrow \mathrm{X}^{2-}+\mathrm{H}_{2} \mathrm{O} $$ The ion formed as a product, \(\mathrm{X}^{2-}\), was shown to have 36 total electrons. What is element \(\mathrm{X}\) ? Propose a name for \(\mathrm{H}_{2} \mathrm{X}\). To completely neutralize a sample of \(\mathrm{H}_{2} \mathrm{X}, 35.6 \mathrm{~mL}\) of \(0.175 \mathrm{M} \mathrm{OH}^{-}\) solution was required. What was the mass of the \(\mathrm{H}_{2} \mathrm{X}\) sample used?

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