Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 125

Bornite \(\left(\mathrm{Cu}_{3} \mathrm{FeS}_{3}\right)\) is a copper ore used in the production of copper. When heated, the following reaction occurs: $$2 \mathrm{Cu}_{3} \mathrm{FeS}_{3}(s)+7 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{Cu}(s)+2 \mathrm{FeO}(s)+6 \mathrm{SO}_{2}(g)$$ If 2.50 metric tons of bornite is reacted with excess \(\mathrm{O}_{2}\) and the process has an \(86.3 \%\) yield of copper, what mass of copper is produced?

Short Answer

Expert verified
The mass of copper produced in this reaction is approximately 1793263 g, or 1.79 metric tons.

Step by step solution

01

Convert mass of bornite to moles

To begin, we must convert the given mass of bornite (2.50 metric tons) to moles. We need to know the molar mass of bornite to do this. 1 metric ton = 1000 kg 2.50 MT of bornite = 2.50 * 1000 kg = 2500 kg Molar mass of \(\mathrm{Cu}_{3} \mathrm{FeS}_{3} = 3(63.5) + 55.85 + 3(32.1) = 228.65 \ g/mol\) Now let's convert mass of bornite to moles: \[ \frac{2500 \ kg}{1} \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol}{228.65 \ g} \approx 10919 \ moles \ of \ \mathrm{Cu}_{3} \mathrm{FeS}_{3} \]
02

Determine moles of copper produced

Use the stoichiometry from the balanced chemical equation to determine moles of copper produced. \[ 10919 \ moles \ of \ \mathrm{Cu}_{3} \mathrm{FeS}_{3} \times \frac{6 \ moles \ of \ Cu}{2 \ moles \ of \ \mathrm{Cu}_{3} \mathrm{FeS}_{3}} =32757 \ moles \ of \ Cu \]
03

Calculate the theoretical mass of copper produced

Now convert the moles of copper to mass. \[ 32757 \ moles \ of \ Cu \times \frac{63.5 \ g}{1 \ mol} \approx 2078035 \ g \]
04

Find the actual mass of copper produced using the yield

We are given an 86.3% yield of copper. Use this percentage to find the actual mass of copper produced. \[ 2078035 \ g \times \frac{86.3\%}{100} \approx 1793263 \ g \] The mass of copper produced in this reaction is approximately 1793263 g, or 1.79 metric tons.

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

Acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\right)\) is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. $$2 \mathrm{C}_{3} \mathrm{H}_{6}(g)+2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ If \(15.0 \mathrm{g}\) \(\mathrm{C}_{3} \mathrm{H}_{6}, 10.0 \mathrm{g} \mathrm{O}_{2},\) and \(5.00 \mathrm{g}\) \(\mathrm{NH}_{3}\) are reacted, what mass of acrylonitrile can be produced, assuming \(100 \%\) yield?

A given sample of a xenon fluoride compound contains molecules of the type \(\mathrm{XeF}_{n},\) where \(n\) is some whole number. Given that \(9.03 \times 10^{20}\) molecules of \(\mathrm{XeF}_{n}\) weigh \(0.368 \mathrm{g},\) determine the value for \(n\) in the formula.

Consider the following unbalanced reaction: $$\mathrm{P}_{4}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{PF}_{3}(g)$$ What mass of \(\mathrm{F}_{2}\) is needed to produce \(120 . \mathrm{g}\) of \(\mathrm{PF}_{3}\) if the reaction has a \(78.1 \%\) yield?

Determine the molecular formulas to which the following empirical formulas and molar masses pertain. a. SNH (188.35 g/mol) c. \(\mathrm{CoC}_{4} \mathrm{O}_{4}(341.94 \mathrm{g} / \mathrm{mol})\) b. \(\mathrm{NPCl}_{2}(347.64 \mathrm{g} / \mathrm{mol})\) d. \(\mathrm{SN}(184.32 \mathrm{g} / \mathrm{mol})\)

When the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). In a certain experiment, \(20.00 \mathrm{g}\) iron metal was reacted with \(11.20 \mathrm{g}\) oxygen gas. After the experiment, the iron was totally consumed, and 3.24 g oxygen gas remained. Calculate the amounts of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) formed in this experiment.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

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