a. Iron (III) oxide reacts with hydrogen gas to form elemental iron and water, according to the balanced equation shown below. How many moles of iron will be formed from the reduction of excess iron (III) oxide by \(0.58\) moles of hydrogen gas? b. When an impure sample containing an unknown amount of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is reacted with excess hydrogen gas, \(0.16\) moles of solid Fe are formed. How many moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) were in the original sample?

Short Answer

Expert verified
a. 0.387 moles of iron will be formed. b. There were 0.08 moles of Fe2O3 in the original sample.

Step by step solution

01

Identify the Reaction and the Balanced Equation

Write down the chemical equation for the reaction and balance it if it is not already balanced. The reaction of iron (III) oxide with hydrogen gas is represented by the equation: \[\mathrm{Fe_2O_3(s) + 3H_2(g) \rightarrow 2Fe(s) + 3H_2O(g)}\]The equation is already balanced.
02

Calculate Moles of Iron Formed from Hydrogen Gas (Part a)

Using the balanced equation, determine the mole ratio between hydrogen gas and iron. According to the equation, 3 moles of hydrogen gas produce 2 moles of iron. With the given 0.58 moles of hydrogen gas, apply the stoichiometric ratio: \[\frac{2 \text{ moles Fe}}{3 \text{ moles H}_2} = \frac{x}{0.58}\] Solve for x to find the number of moles of iron.
03

Solve for Moles of Iron (Part a)

To find the number of moles of iron from the stoichiometric ratio, calculate x as follows: \[x = \frac{2}{3} \times 0.58\]\[x = 0.387 \text{ moles of Fe}\]
04

Use Product-to-Reactant Mole Ratio (Part b)

To find moles of iron (III) oxide from the produced iron, use the reverse stoichiometric ratio from iron to iron (III) oxide from the balanced equation (2 mol Fe to 1 mol Fe2O3). Given that 0.16 moles of Fe are formed, apply the stoichiometric ratio: \[\frac{1 \text{ mole Fe}_2\text{O}_3}{2 \text{ moles Fe}} = \frac{x}{0.16}\] Solve for x to find the number of moles of iron (III) oxide in the original sample.
05

Calculate Moles of Iron (III) Oxide (Part b)

To find moles of iron (III) oxide from iron, calculate x as follows: \[x = \frac{1}{2} \times 0.16\]\[x = 0.08 \text{ moles of Fe}_2\text{O}_3\]

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 Vaia!

Key Concepts

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

Chemical Equations
Chemical equations are a formal representation of chemical reactions, depicting the substances involved and their transformation. They are composed of chemical formulas that represent the reactants and products separated by an arrow indicating the direction of the reaction. A key characteristic of a balanced chemical equation is that the number of atoms for each element must be equal on both sides, ensuring that matter is conserved.

For instance, the equation provided in the exercise, \[\mathrm{Fe_2O_3(s) + 3H_2(g) \rightarrow 2Fe(s) + 3H_2O(g)}\] clearly shows the reactant iron (III) oxide (\(\mathrm{Fe_2O_3}\)) and hydrogen gas (\(\mathrm{H_2}\)) transforming into iron (\(\mathrm{Fe}\)) and water (\(\mathrm{H_2O}\)) as products. This equation is deemed balanced because the number of atoms for each element—iron, oxygen, and hydrogen—remains unchanged from the reactants side to the products side.
Mole Ratio
The mole ratio, derived from the coefficients of a balanced chemical equation, is critical for stoichiometric calculations. It establishes a proportional relationship between the quantities of reactants and products involved in a chemical reaction.

In the provided exercise, the mole ratio between hydrogen gas and iron is derived as follows: for every 3 moles of hydrogen gas used, 2 moles of iron are produced. Mathematically, this ratio is represented as \(\frac{2 \text{ moles Fe}}{3 \text{ moles H}_2}\). It's important to understand that these ratios are based on the coefficients in the balanced equation, and they allow us to convert between moles of different substances involved in the reaction.
Stoichiometric Calculations
Stoichiometric calculations enable us to use the mole ratio from the balanced chemical equation to determine the quantity of one substance, given the quantity of another. These calculations require a step-by-step approach that typically includes: writing out the balanced equation, determining the mole ratio, and then using this ratio to calculate the unknown quantity.

For example, to calculate moles of iron formed from hydrogen gas in part (a) of the exercise, we multiply the given moles of hydrogen gas by the mole ratio, which derives from the balanced equation. To find the moles of iron \(\mathrm{(Fe)}\), the calculation is as follows: \[x = \frac{2}{3} \times 0.58\]\[x = 0.387 \text{ moles of Fe}\]. This process is standard in stoichiometry, and it's pivotal to ensure that all steps and calculations adhere to the mole ratios established by the balanced equation.
Chemical Reaction
A chemical reaction is a process in which reactants are transformed into products through the breaking and forming of chemical bonds. It involves a change in the chemical substances and is often accompanied by visual changes like color change, temperature change, gas formation, or precipitate formation. The reaction must be depicted by a balanced chemical equation that reflects the conservation of mass.

In the given exercise, we see a reduction-oxidation (redox) reaction taking place, where iron (III) oxide reacts with hydrogen gas—a reduction of iron oxide accompanied by the oxidation of hydrogen. This reaction not only illustrates the conversion of substances but also represents the practical application of stoichiometry in predicting the outcome of a chemical reaction, which is quintessential in fields like chemistry and engineering.

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

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.

Sign-up for free