/*! 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 125 Bornite \(\left(\mathrm{Cu}_{3} ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Chemical Reactions
Chemical reactions are processes where substances, called reactants, change to form different substances, known as products. In our example, the chemical reaction involves bornite \((\mathrm{Cu}_{3} \mathrm{FeS}_{3})\) and oxygen \((\mathrm{O}_{2})\) as reactants, which react to produce copper \((\mathrm{Cu})\), iron(II) oxide \((\mathrm{FeO})\), and sulfur dioxide \((\mathrm{SO}_{2})\).
Understanding these reactions requires us to interpret the balanced chemical equation provided. This balance is important because it shows that mass is conserved during the reaction, with the same number of atoms on both sides of the equation.
Balanced equations also help us understand stoichiometry, as they provide the mole ratio between reactants and products. In the bornite reaction, for each 2 moles of bornite, 6 moles of copper are produced. Knowing this relation is key to finding out how much copper we can potentially produce.
Copper Production
Copper production from ores like bornite involves extracting the metal from its mineral state through a series of chemical processes. Bornite is one of the many copper ores used due to its copper-rich nature.
When bornite is heated and reacts with oxygen, copper is one of the primary products, extracted from the mineral and separated from other elements like iron and sulfur. This process forms a core part of metallurgical chemistry where raw ores undergo transformations to yield metals like copper for industrial use.
In the given reaction, bornite is reacted in excess oxygen conditions, which means that oxygen is plentiful and helps ensure complete reaction of all available bornite to maximize copper yield. This detail indicates efficient production, as unreacted oxygen does not limit the amount of product formed, focusing instead on the complete utilization of the ore.
Theoretical Yield
Theoretical yield is the maximum amount of product that can be produced in a chemical reaction under ideal conditions. In practice, it's calculated based on the stoichiometry of the balanced equation alongside the initial amount of reactants.
For the bornite reaction, once we've converted the reactant mass to moles, we use the stoichiometric ratios from the chemical equation to find the expected moles of copper. From there, we calculate the mass that these moles of copper would correspond to, giving us the theoretical yield.
The calculated mass of 2078035 g represents this maximum potential output of copper if everything reacts perfectly. However, in real-world applications, actual production may be less due to inefficiencies or side reactions, which is why we also consider the percent yield.
Molar Mass Calculation
Molar mass is a crucial concept in stoichiometry and chemistry, as it converts between the mass of a substance and the amount in moles, enabling the use of balanced equations.
Calculating it involves summing the atomic masses of all elements in a compound. For bornite \((\mathrm{Cu}_{3} \mathrm{FeS}_{3})\), the molar mass is calculated as follows:
  • Copper (\(\mathrm{Cu}\)): 3 atoms \(\times 63.5 \ \text{g/mol} = 190.5 \ \text{g/mol}\)
  • Iron (\(\mathrm{Fe}\)): 1 atom \(\times 55.85 \ \text{g/mol} = 55.85 \ \text{g/mol}\)
  • Sulfur (\(\mathrm{S}\)): 3 atoms \(\times 32.1 \ \text{g/mol} = 96.3 \ \text{g/mol}\)
Adding these, we find the molar mass of bornite to be 228.65 g/mol. This figure then helps convert the mass of bornite available into moles so we can use stoichiometric relationships to predict product quantities from 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

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.

Many cereals are made with high moisture content so that the cereal can be formed into various shapes before it is dried. A cereal product containing \(58 \% \mathrm{H}_{2} \mathrm{O}\) by mass is produced at the rate of \(1000 .\) kg/h. What mass of water must be evaporated per hour if the final product contains only \(20 . \%\) water?

A \(2.077-\mathrm{g}\) sample of an element, which has an atomic mass between 40 and \(55,\) reacts with oxygen to form 3.708 g of an oxide. Determine the formula of the oxide (and identify the element).

Ammonia reacts with \(\mathrm{O}_{2}\) to form either \(\mathrm{NO}(g)\) or \(\mathrm{NO}_{2}(g)\) according to these unbalanced equations: $$\begin{array}{l}\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\\\\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\end{array}$$ In a certain experiment 2.00 moles of \(\mathrm{NH}_{3}(g)\) and 10.00 moles of \(\mathbf{O}_{2}(g)\) are contained in a closed flask. After the reaction is complete, 6.75 moles of \(\mathbf{O}_{2}(g)\) remains. Calculate the number of moles of \(\mathrm{NO}(g)\) in the product mixture: (Hint: You cannot do this problem by adding the balanced equations because you cannot assume that the two reactions will occur with equal probability.)

A compound with molar mass \(180.1 \mathrm{g} / \mathrm{mol}\) has the following composition by mass: $$\begin{array}{|ll|}\hline C & 40.0 \% \\\H & 6.70 \% \\\O & 53.3 \% \\\\\hline\end{array}$$ Determine the empirical and molecular formulas of the compound.

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.