Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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%.

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 Vaia!

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.

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

A \(2.20-\) g sample of an unknown acid (empirical formula \(=\) \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{3} )\) is dissolved in 1.0 \(\mathrm{L}\) of water. A titration required 25.0 mL of 0.500 M NaOH to react completely with all the acid present. Assuming the unknown acid has one acidic proton per molecule, what is the molecular formula of the unknown acid?

Assign oxidation states for all atoms in each of the following compounds a. \({KMnO}_{4} \quad\quad\quad f. {Fe}_{3} {O}_{4}\) b. \({NiO}_{2} \quad\quad\quad\quad g. {XeOF}_{4}\) c. \({Na}_{4} {Fe}({OH})_{6} \quad h. {SF}_{4}\) d. \({NH}_{4} {h}_{2} {HPO}_{4} \quad i. {CO}\) e. \({P}_{4} {O}_{6} \quad\quad\quad\quad\quad j. {C}_{6} {H}_{12} {O}_{6}\)

A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are 50.0 mL of 0.100 M hydrochloric acid, 100.0 mL of 0.200 M of nitric acid, 500.0 mL of 0.0100 M calcium hydroxide, and 200.0 mL of 0.100 M rubidium hydroxide. Did the acids and bases exactly neutralize each other? If not, calculate the concentration of excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in solution.

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?

Chlorisondamine chloride \(\left(\mathrm{C}_{14} \mathrm{H}_{20} \mathrm{Cl}_{6} \mathrm{N}_{2}\right)\) is a drug used in the treatment of hypertension. A 1.28-g sample of a medication containing the drug was treated to destroy the organic material and to release all the chlorine as chloride ion. When the filtered solution containing chloride ion was treated with an excess of silver nitrate, 0.104 g silver chloride was recovered. Calculate the mass percent of chlorisondamine chloride in the medication, assuming the drug is the only source of chloride.

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.

Sign-up for free