/*! 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 65 Aluminum sulfide reacts with wat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Aluminum sulfide reacts with water to form aluminum hydroxide and hydrogen sulfide. (a) Write the balanced chemical equation for this reaction. (b) How many grams of aluminum hydroxide are obtained from \(14.2 \mathrm{~g}\) of aluminum sulfide?

Short Answer

Expert verified
(a) The balanced chemical equation for the reaction is: Al2S3 (s) + 6H2O (l) -> 2Al(OH)3 (aq) + 3H2S (g) (b) Approximately 14.77 g of aluminum hydroxide are obtained from 14.2 g of aluminum sulfide.

Step by step solution

01

Write the balanced chemical equation

The unbalanced chemical equation for the reaction can be written as: Al2S3 (s) + H2O (l) -> Al(OH)3 (aq) + H2S (g) To balance the equation, we need to make sure there are the same number of each element on both the reactant side and the product side. After balancing, the equation becomes: Al2S3 (s) + 6H2O (l) -> 2Al(OH)3 (aq) + 3H2S (g)
02

Calculate the moles of aluminum sulfide

To calculate the moles of aluminum sulfide, we can use the given mass (14.2 g) and molar mass of aluminum sulfide, which is approximately 150 g/mol (26.98 g/mol for aluminum and 32.07 g/mol for sulfur). Moles of Al2S3 = (14.2 g) / (150 g/mol) = 0.0947 mol
03

Use the mole ratio to find the moles of aluminum hydroxide produced

From the balanced equation, we can see that 1 mole of Al2S3 produces 2 moles of Al(OH)3. So, with the available moles of aluminum sulfide: Moles of Al(OH)3 = Moles of Al2S3 × (2 moles of Al(OH)3 / 1 mole of Al2S3) = 0.0947 mol × 2 = 0.1894 mol
04

Calculate the mass of aluminum hydroxide produced

To find the mass of aluminum hydroxide produced, we can use the moles calculated in the previous step and the molar mass of aluminum hydroxide. The molar mass of Al(OH)3 is approximately 78 g/mol (26.98 g/mol for aluminum and 3 × 16 g/mol for oxygen plus 3 × 1 g/mol for hydrogen): Mass of Al(OH)3 = Moles of Al(OH)3 × Molar mass of Al(OH)3 Mass of Al(OH)3 = (0.1894 mol) × (78 g/mol) = 14.77 g So, from 14.2 g of aluminum sulfide, approximately 14.77 g of aluminum hydroxide are produced.

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!

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

Hydrofluoric acid, \(\mathrm{HF}(a q),\) cannot be stored in glass bottles because compounds called silicates in the glass are attacked by the \(\mathrm{HF}(a q)\). Sodium silicate \(\left(\mathrm{Na}_{2} \mathrm{SiO}_{3}\right)\), for example, reacts as follows: $$ \begin{aligned} \mathrm{Na}_{2} \mathrm{SiO}_{3}(s)+8 \mathrm{HF}(a q) & \longrightarrow \\ \mathrm{H}_{2} \mathrm{SiF}_{6}(a q) &+2 \mathrm{NaF}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ (a) How many moles of HF are needed to react with 0.300 \(\mathrm{mol}\) of \(\mathrm{Na}_{2} \mathrm{SiO}_{3} ?\) (b) How many grams of NaF form when \(0.500 \mathrm{~mol}\) of HF reacts with excess \(\mathrm{Na}_{2} \mathrm{SiO}_{3} ?\) (c) How many grams of \(\mathrm{Na}_{2} \mathrm{SiO}_{3}\) can react with \(0.800 \mathrm{~g}\) of HF?

Epsom salts, a strong laxative used in veterinary medicine, is a hydrate, which means that a certain number of water molecules are included in the solid structure. The formula for Epsom salts can be written as \(\mathrm{MgSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O},\) where \(x\) indicates the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{MgSO}_{4}\). When \(5.061 \mathrm{~g}\) of this hydrate is heated to \(250{ }^{\circ} \mathrm{C},\) all the water of hydration is lost, leaving \(2.472 \mathrm{~g}\) of \(\mathrm{MgSO}_{4}\). What is the value of \(x ?\)

The source of oxygen that drives the internal combustion engine in an automobile is air. Air is a mixture of gases, principally \(\mathrm{N}_{2}(\sim 79 \%)\) and \(\mathrm{O}_{2}(\sim 20 \%) .\) In the cylinder of an automobile engine, nitrogen can react with oxygen to produce nitric oxide gas, NO. As \(\mathrm{NO}\) is emitted from the tailpipe of the car, it can react with more oxygen to produce nitrogen dioxide gas. (a) Write balanced chemical equations for both reactions. (b) Both nitric oxide and nitrogen dioxide are pollutants that can lead to acid rain and global warming; collectively, they are called " \(\mathrm{NO}_{\mathrm{x}}^{\prime \prime}\) gases. In \(2007,\) the United States emitted an estimated 22 million tons of nitrogen dioxide into the atmosphere. How many grams of nitrogen dioxide is this? (c) The production of \(\mathrm{NO}_{\mathrm{x}}\) gases is an unwanted side reaction of the main engine combustion process that turns octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), into \(\mathrm{CO}_{2}\) and water. If \(85 \%\) of the oxygen in an engine is used to combust octane and the remainder used to produce nitrogen dioxide, calculate how many grams of nitrogen dioxide would be produced during the combustion of 500 grams of octane.

The koala dines exclusively on eucalyptus leaves. Its digestive system detoxifies the eucalyptus oil, a poison to other animals. The chief constituent in eucalyptus oil is a substance called eucalyptol, which contains \(77.87 \% \mathrm{C}, 11.76 \% \mathrm{H},\) and the remainder O. (a) What is the empirical formula for this substance? (b) A mass spectrum of eucalyptol shows a peak at about 154 amu. What is the molecular formula of the substance?

Aluminum hydroxide reacts with sulfuric acid as follows: $$ 2 \mathrm{Al}(\mathrm{OH})_{3}(s)+3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ Which is the limiting reactant when \(0.500 \mathrm{~mol} \mathrm{Al}(\mathrm{OH})_{3}\) and \(0.500 \mathrm{~mol} \mathrm{H}_{2} \mathrm{SO}_{4}\) are allowed to react? How many moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) can form under these conditions? How many moles of the excess reactant remain after the completion of the reaction?

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.