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Chapter 4: Problem 110

A mixture contains only \(\mathrm{NaCl}\) and \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) . A \(1.45-\mathrm{g}\) sample of the mixture is dissolved in water and an excess of NaOH is added, producing a precipitate of \(\mathrm{Al}(\mathrm{OH})_{3}\) . The precipitate is filtered, dried, and weighed. The mass of the precipitate is 0.107 g. What is the mass percent of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in the sample?

Short Answer

Expert verified
The mass percent of $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ in the mixture is 32.34%.

Step by step solution

01

Find the moles of Al(OH)₃

First, we need to find the number of moles of Al(OH)₃ precipitate formed. For this, we can use the mass of the precipitate and the molar mass of Al(OH)₃. The molar mass of Al(OH)₃ is: M_Al(OH)₃ = 27.0 g/mol (Al) + 3*(16.0 g/mol (O) + 1.0 g/mol (H)) = 27.0 + 3*(17.0) = 27.0 + 51.0 = 78.0 g/mol Now, we can find the moles of Al(OH)₃ formed: moles_Al(OH)₃ = mass_Al(OH)₃ / M_Al(OH)₃ = 0.107 g / 78.0 g/mol = 0.00137 mol
02

Find the moles of Al₂(SO₄)₃

Since every mole of Al(OH)₃ is formed from one mole of Al₂(SO₄)₃ in the reaction, the number of moles of Al₂(SO₄)₃ in the sample is equal to the moles of Al(OH)₃: moles_Al₂(SO₄)₃ = moles_Al(OH)₃ = 0.00137 mol
03

Calculate the mass of Al₂(SO₄)₃

Now we can find the mass of Al₂(SO₄)₃ in the sample using the number of moles and molar mass of Al₂(SO₄)₃. The molar mass of Al₂(SO₄)₃ is: M_Al₂(SO₄)₃ = 2*27.0 g/mol (Al) + 3*(4*16.0 g/mol (O) + 32.1 g/mol (S)) = 54.0 + 3*(64.0 + 32.1) = 54.0 + 3*(96.1) = 54.0 + 288.3 = 342.3 g/mol Now, let's calculate the mass: mass_Al₂(SO₄)₃ = moles_Al₂(SO₄)₃ * M_Al₂(SO₄)₃ = 0.00137 mol * 342.3 g/mol = 0.469 g
04

Find the mass percent of Al₂(SO₄)₃ in the mixture

Finally, we can calculate the mass percent of Al₂(SO₄)₃ in the mixture using the mass of Al₂(SO₄)₃ and the mass of the sample: mass_percent_Al₂(SO₄)₃ = (mass_Al₂(SO₄)₃ / mass_sample) * 100% = (0.469 g / 1.45 g) * 100% = 32.34% The mass percent of Al₂(SO₄)₃ in the mixture is 32.34%.

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Key Concepts

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

Molar Mass
Understanding the molar mass is crucial when dealing with chemical reactions and stoichiometry. The molar mass of a compound is the sum of the masses of all the atoms in a molecule. This is expressed in grams per mole (g/mol). To find the molar mass of a chemical compound, you need to:
  • Identify the atomic masses of each atom using the periodic table.
  • Multiply the atomic mass of each element by the number of times it appears in the compound.
  • Add all these values together to get the total molar mass.
For instance, in our exercise, the molar mass of aluminum hydroxide \(\mathrm{Al(OH)_{3}}\) is calculated as 78.0 g/mol. This helps us determine the number of moles when we know the mass of the compound.
Precipitation Reaction
Precipitation reactions are processes where two soluble salts in solution react to form one or more insoluble substances, known as precipitates. These reactions are common in solutions, and the resulting precipitate is often easily separated from the mixture.
In our exercise, an excess of sodium hydroxide \(\mathrm{NaOH}\) was added to a solution containing aluminum sulfate \(\mathrm{Al_{2}(SO_{4})_{3}}\). This led to the formation of aluminum hydroxide \(\mathrm{Al(OH)_{3}}\) as a precipitate, which was then filtered, dried, and weighed. Such reactions are vital for various applications, including:
  • Purifying chemicals by removing dissolved impurities.
  • Identifying ions present in a solution through qualitative analysis.
  • Producing solids with specific properties for industrial uses.
Mass Percent Composition
Mass percent composition is a way to express the concentration of a component in a mixture or a compound. It is particularly useful in chemistry for determining the proportion of each element or compound present in a mixture. To calculate the mass percent of a substance in a mixture:
  • First, determine the mass of the specific component in the mixture.
  • Divide this by the total mass of the mixture or sample.
  • Multiply by 100 to convert to a percentage.
In the given problem, we calculated the mass of \(\mathrm{Al_{2}(SO_{4})_{3}}\) and used it to find its mass percent in the mixture. Determining mass composition helps chemists understand how elements and compounds are distributed, which is crucial for formulating substances and balancing chemical reactions.

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Most popular questions from this chapter

Specify which of the following are oxidation–reduction reactions, and identify the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced. a. \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \rightarrow 2 \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)\) b. \(\mathrm{HCl}(g)+\mathrm{NH}_{3}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) c. \(\mathrm{SiCl}_{4}(i)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 4 \mathrm{HCl}(a q)+\mathrm{SiO}_{2}(s)\) d. \(\mathrm{SiCl}_{4}(l)+2 \mathrm{Mg}(s) \rightarrow 2 \mathrm{MgCl}_{2}(s)+\mathrm{Si}(s)\) e. \(\mathrm{Al}(\mathrm{OH})_{4}-(a q) \rightarrow \mathrm{AlO}_{2}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

The unknown acid \(\mathrm{H}_{2} \mathrm{X}\) can be neutralized completely by \(\mathrm{OH}^{-}\) according to the following (unbalanced) equation: $$\mathrm{H}_{2} \mathrm{X}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{X}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(i) $$ The ion formed as a product, \(X^{2-},\) was shown to have 36 total electrons. What is element 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?

Balance the following oxidation–reduction reactions that occur in basic solution. a. \(\mathrm{Al}(s)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)+\mathrm{Al}(\mathrm{OH})_{4}-(a q)\) b. \(\mathrm{Cl}_{2}(g) \rightarrow \mathrm{Cl}^{-}(a q)+\mathrm{OCl}^{-}(a q)\) c. \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{AlO}_{2}^{-}(a q)\)

Triiodide ions are generated in solution by the following (unbalanced) reaction in acidic solution: $$\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{3}^{-}(a q)$$ Triodide ion concentration is determined by titration with a sodium thiosulfate \(\left(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\right)\) solution. The products are iodide ion and tetrathionate ion \(\left(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\right)\) a. Balance the equation for the reaction of \(\mathrm{IO}_{3}^{-}\) with \(\mathrm{I}^{-}\) ions. b. A sample of 0.6013 g of potassium iodate was dissolved in water. Hydrochloric acid and solid potassium iodide were then added. What is the minimum mass of solid KI and the minimum volume of 3.00 M HCl required to convert all of the 1 \(\mathrm{O}_{3}^{-}\) ions to \(\mathrm{I}_{3}^{-}\) ions? c. Write and balance the equation for the reaction of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) with \(\mathrm{I}_{3}-\) in acidic solution. d. A 25.00 -mL sample of a 0.0100\(M\) solution of \(\mathrm{KIO}_{3}\) is reacted with an excess of \(\mathrm{KL}\) . It requires 32.04 \(\mathrm{mL}\) of \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) solution to titrate the \(\mathrm{I}_{3}^{-}\) ions present. What is the molarity of the \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) solution? e. How would you prepare 500.0 \(\mathrm{mL}\) of the KIO\(_{3}\)solution in part d using solid \(\mathrm{KIO}_{3} ?\)

What mass of iron(III) hydroxide precipitate can be produced by reacting 75.0 mL of 0.105 M iron(III) nitrate with 125 mL of 0.150 M sodium hydroxide?
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