Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 129

Aluminum reacts with bromine, producing aluminum bromide: $$ 2 \mathrm{Al}(s)+3 \mathrm{Br}_{2}(l) \rightarrow 2 \mathrm{AlBr}_{3}(s) $$ In a certain experiment, 20.0 \(\mathrm{mL}\) of bromine (density \(=\) 3.10 \(\mathrm{g} / \mathrm{mL}\) ) was reacted with excess aluminum to yield 50.3 \(\mathrm{g}\) of aluminum bromide. What is the percent yield for this experiment?

Short Answer

Expert verified
The percent yield for this experiment is approximately 73.0%.

Step by step solution

01

Calculate the mass of bromine used

To calculate the mass of bromine used, we will multiply the given volume (20.0 mL) by the density (3.10 g/mL): Mass of bromine = (20.0 mL) * (Density) \(=20.0\,\mathrm{mL}\times3.10\,\frac{\mathrm{g}}{\mathrm{mL}}=62.0\,\mathrm{g}\) So, the mass of bromine used in the reaction is 62.0 g.
02

Calculate the moles of bromine

We have to convert the mass of bromine into moles using the molar mass of bromine. The molar mass of Br2 is approximately 159.8 g/mol. Moles of bromine = mass / molar mass \(\frac{62.0\,\mathrm{g}}{159.8\,\frac{\mathrm{g}}{\mathrm{mol}}}\approx0.388\,\mathrm{mol}\) So we have 0.388 moles of bromine.
03

Calculate the moles of aluminum bromide that can theoretically be produced

Using the reaction stoichiometry, 3 moles of bromine react with 2 moles of aluminum to produce 2 moles of aluminum bromide. Thus, to calculate the moles of aluminum bromide that can theoretically be produced, we can use a mole-to-mole ratio. Moles of aluminum bromide = Moles of bromine * (2 / 3) \(= 0.388\,\mathrm{mol}\times \frac{2}{3}\approx 0.2587\,\mathrm{mol}\) So, approximately 0.2587 moles of aluminum bromide can theoretically be produced.
04

Calculate the theoretical mass of aluminum bromide

We have to convert the moles of aluminum bromide back into mass using the molar mass of aluminum bromide, which is approximately 266.7 g/mol. Theoretical mass of aluminum bromide = Moles of aluminum bromide * molar mass \(= 0.2587\,\mathrm{mol}\times 266.7\,\frac{\mathrm{g}}{\mathrm{mol}} \approx 68.9\,\mathrm{g}\) So, the theoretical mass of aluminum bromide is 68.9 g.
05

Calculate the percent yield of the reaction

Now that we have the actual mass of aluminum bromide really produced (50.3 g) and the theoretical mass of aluminum bromide that could be produced (68.9 g), we can calculate the percent yield. Percent yield = (Actual mass / Theoretical mass) * 100 \(\frac{50.3\,\mathrm{g}}{68.9\,\mathrm{g}} \times100\approx73.0\%\) So, the percent yield for this experiment is approximately 73.0%.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Maleic acid is an organic compound composed of \(41.39 \% \mathrm{C},\) 3.47$\% \mathrm{H}$ , and the rest oxygen. If 0.129 mole of maleic acid has a mass of 15.0 \(\mathrm{g}\) , what are the empirical and molecular formulas of maleic acid?

Para-cresol, a substance used as a disinfectant and in the manufacture of several herbicides, is a molecule that contains the elements carbon, hydrogen, and oxygen. Complete combustion of a \(0.345-\mathrm{g}\) sample of \(p\) -cresol produced 0.983 g carbon dioxide and 0.230 \(\mathrm{g}\) water. Determine the empirical formula for \(p\) -cresol.

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ a. What is the maximum mass of ammonia that can be produced from a mixture of \(1.00 \times 10^{3} \mathrm{g} \mathrm{N}_{2}\) and $5.00 \times 10^{2} \mathrm{g} \mathrm{H}_{2} ?$ b. What mass of which starting material would remain unreacted?

Silicon is produced for the chemical and electronics industries by the following reactions. Give the balanced equation for each reaction. a. \(\mathrm{SiO}_{2}(s)+\mathrm{C}(s)\) $\frac{\text { Electric }}{\text { are furnace }}\( \)\mathrm{Si}(s)+\mathrm{CO}(g)$ b. Liquid silicon tetrachloride is reacted with very pure solid magnesium, producing solid silicon and solid magnesium chloride. c. $\mathrm{Na}_{2} \mathrm{SiF}_{6}(s)+\mathrm{Na}(s) \rightarrow \mathrm{Si}(s)+\mathrm{NaF}(s)$

Which of the following statements about chemical equations is(are) true? a. When balancing a chemical equation, you can never change the coefficient in front of any chemical formula. b. The coefficients in a balanced chemical equation refer to the number of grams of reactants and products. c. In a chemical equation, the reactants are on the right and the products are on the left. d. When balancing a chemical equation, you can never change the subscripts of any chemical formula. e. In chemical reactions, matter is neither created nor destroyed so a chemical equation must have the same number of atoms on both sides of the equation.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation
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