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Iron ore is converted to iron metal in a reaction with carbon. $$ 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{C}(\mathrm{s}) \rightarrow 4 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g}) $$ If 6.2 mol of \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})\) is used, what amount of \(\mathrm{C}(\mathrm{s})\) is needed, and what amounts of Fe and \(\mathrm{CO}_{2}\) are produced?

Short Answer

Expert verified
9.3 mol C, 12.4 mol Fe, 9.3 mol CO2

Step by step solution

01

Write down the balanced equation

The chemical reaction is given as \( 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s}) + 3 \mathrm{C}(\mathrm{s}) \rightarrow 4 \mathrm{Fe}(\mathrm{s}) + 3 \mathrm{CO}_{2}(\mathrm{g}) \). This equation is balanced, meaning the number of each type of atom is equal on both sides of the equation.
02

Analyze the stoichiometry of reactants

From the balanced equation, 2 moles of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \) requires 3 moles of \( \mathrm{C} \). Use this relationship to determine the amount of \( \mathrm{C} \) needed.
03

Calculate moles of carbon needed

For 2 moles of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \), 3 moles of \( \mathrm{C} \) are needed. Therefore, for 6.2 moles of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \), the calculation is: \[ 6.2 \text{ moles of } \mathrm{Fe}_{2} \mathrm{O}_{3} \times \frac{3 \text{ moles of } \mathrm{C}}{2 \text{ moles of } \mathrm{Fe}_{2} \mathrm{O}_{3}} = 9.3 \text{ moles of } \mathrm{C} \].
04

Determine products using stoichiometry

The balanced equation indicates that 4 moles of \( \mathrm{Fe} \) and 3 moles of \( \mathrm{CO}_{2} \) are produced per 2 moles of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \).
05

Calculate moles of Fe produced

For 6.2 moles of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \), calculate \( \mathrm{Fe} \) produced: \[ 6.2 \text{ moles of } \mathrm{Fe}_{2} \mathrm{O}_{3} \times \frac{4 \text{ moles of } \mathrm{Fe}}{2 \text{ moles of } \mathrm{Fe}_{2} \mathrm{O}_{3}} = 12.4 \text{ moles of } \mathrm{Fe} \].
06

Calculate moles of CO2 produced

Similarly, calculate \( \mathrm{CO}_{2} \) produced: \[ 6.2 \text{ moles of } \mathrm{Fe}_{2} \mathrm{O}_{3} \times \frac{3 \text{ moles of } \mathrm{CO}_{2}}{2 \text{ moles of } \mathrm{Fe}_{2} \mathrm{O}_{3}} = 9.3 \text{ moles of } \mathrm{CO}_{2} \].

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Key Concepts

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

Balanced chemical equation
A balanced chemical equation is essential in understanding chemical reactions. It accurately represents the quantities of reactants and products involved in a chemical reaction. In this context, **balancing** a chemical equation ensures that the number of atoms of each element is the same on both sides of the equation. This preserves the law of conservation of mass.

In the provided exercise, the chemical equation given is:
  • \( 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s}) + 3 \mathrm{C}(\mathrm{s}) \rightarrow 4 \mathrm{Fe}(\mathrm{s}) + 3 \mathrm{CO}_{2}(\mathrm{g}) \)
This equation is balanced. For every element like iron (Fe), carbon (C), and oxygen (O), the atoms on the reactant side equal the atoms on the product side:
  • Iron (Fe): 4 in \(2 \mathrm{Fe}_{2} \mathrm{O}_{3}\) results in 4 in \(4 \mathrm{Fe}\).
  • Carbon (C): 3 in \(3 \mathrm{C}\) results in 3 in \(3 \mathrm{CO}_{2}\).
  • Oxygen (O): 6 in \(2 \mathrm{Fe}_{2} \mathrm{O}_{3}\) results in 6 in \(3 \mathrm{CO}_{2}\).
Balancing can sometimes be tricky, but keeping track of each type of atom can make it more straightforward.
Mole concept
The mole concept is a fundamental principle in chemistry. It's a way to express amounts of a substance that allow chemists to work with atoms and molecules more conveniently. **One mole** contains Avogadro's number of particles, which is \(6.022 \times 10^{23}\) entities. This makes it useful for converting between atoms/molecules and grams.

In stoichiometry, the mole concept helps to solve problems by relating mass and quantity of a particular substance involved in a chemical equation. For instance, from our balanced equation, we understand that:
  • 2 moles of \( \mathrm{Fe}_{2} \mathrm{O}_{3}\) use 3 moles of carbon.
  • It also produces 4 moles of iron and 3 moles of \( \mathrm{CO}_{2}\).
Using this information, if we start with 6.2 moles of \( \mathrm{Fe}_{2} \mathrm{O}_{3}\), we utilize moles of carbon based on these proportions. Simply put, the mole concept serves as a bridge between the microscopic world of atoms and the macroscopic observable quantities we manipulate in the lab.
Chemical reactions
Chemical reactions are processes in which substances combine or change to form new substances. **Reactants** are the starting substances and **products** are the substances formed by the reaction. Understanding the nature of chemical reactions involves comprehending how these transformations occur.

In the example given, iron oxide (\( \mathrm{Fe}_{2} \mathrm{O}_{3}\)) reacts with carbon (\( \mathrm{C}\)) to produce iron (\( \mathrm{Fe}\)) and carbon dioxide (\( \mathrm{CO}_{2}\)). This transformation involves breaking bonds in the reactants and forming new bonds to form the products. Here’s what happens:
  • The oxygen atoms bonded with iron in \( \mathrm{Fe}_{2} \mathrm{O}_{3}\) are captured by carbon to form \( \mathrm{CO}_{2}\).
  • This frees the iron atoms as elemental iron, \( \mathrm{Fe}\).
Chemical reactions can release or absorb energy, typically in the form of heat. This specific reaction is exothermic as it involves the reduction of an oxide by carbon. Understanding these processes is key in fields like metallurgy and environmental science, where control over such reactions is crucial.

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Most popular questions from this chapter

ATOM ECONOMY: Ethylene oxide, \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O},\) is an important industrial chemical las it is the starting place to make such important chemicals as ethylene glycol (antifreeze) and various polymers \(1 .\) One way to make the compound is called the "chlorohydrin route." $$ \mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{Cl}_{2}+\mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}+\mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O} $$ Another route is the modern catalytic reaction. $$ \mathrm{C}_{2} \mathrm{H}_{4}+1 / 2 \mathrm{O}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O} $$ (a) Calculate the \(\%\) atom economy for the production of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) in each of these reactions. Which is the more efficient method? (b) What is the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) if \(867 \mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is used to synthesize \(762 \mathrm{g}\) of the product by the catalytic reaction?

Your body deals with excess nitrogen by excreting it in the form of urea, \(\mathrm{NH}_{2} \mathrm{CONH}_{2}\). The reaction producing it is the combination of arginine \(\left(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{N}_{4} \mathrm{O}_{2}\right)\) with water to give urea and ornithine \(\left(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{N}_{2} \mathrm{O}_{2}\right)\) $$ \mathrm{C}_{6} \mathrm{H}_{14} \mathrm{N}_{4} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}+\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{N}_{2} \mathrm{O}_{2} $$ arginine ornithine If you excrete 95 mg of urea, what mass of arginine must have been used? What mass of ornithine must have been produced?

A Copper metal can be prepared by roasting copper ore, which can contain cuprite \(\left(\mathrm{Cu}_{2} \mathrm{S}\right)\) and copper (11) sulfide. $$ \begin{aligned} \mathrm{Cu}_{2} \mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(g) & \rightarrow 2 \mathrm{Cu}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{g}) \\ \mathrm{CuS}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) & \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{g}) \end{aligned} $$ Suppose an ore sample contains \(11.0 \%\) impurity in addition to a mixture of CuS and Cu \(_{2} \mathrm{S}\). Heating \(100.0 \mathrm{g}\) of the mixture produces \(75.4 \mathrm{g}\) of copper metal with a purity of \(89.5 \% .\) What is the weight percent of CuS in the ore? The weight percent of \(\mathrm{Cu}_{2} \mathrm{S} ?\)

You have \(250 .\) mL of \(0.136 \mathrm{M}\) HCl. Using a volumetric pipet, you take \(25.00 \mathrm{mL}\) of that solution and dilute it to \(100.00 \mathrm{mL}\) in a volumetric flask. Now you take \(10.00 \mathrm{mL}\) of that solution, using a volumetric pipet, and dilute it to \(100.00 \mathrm{mL}\) in a volumetric flask. What is the concentration of hydrochloric acid in the final solution?

The balanced equation for the reduction of iron ore to the metal using CO is $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g}) $$ (a) What is the maximum mass of iron, in grams, that can be obtained from \(454 \mathrm{g}(1.00 \mathrm{lb})\) of iron(III) oxide? (b) What mass of \(\mathrm{CO}\) is required to react with \(454 \mathrm{g}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3} ?\)

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