/*! 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 9 Chromium metal reacts with oxyge... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 9

Chromium metal reacts with oxygen to give chromium(III) oxide, \(\mathrm{Cr}_{2} \mathrm{O}_{3}\) (a) Write a balanced equation for the reaction. (b) What mass (in grams) of \(\mathrm{Cr}_{2} \mathrm{O}_{3}\) is produced if \(0.175 \mathrm{g}\) of chromium metal is converted completely to the oxide? (c) What mass of \(\mathrm{O}_{2}\) (in grams) is required for the reaction?

Short Answer

Expert verified
(a) Balanced equation: \(4\text{Cr} + 3\text{O}_2 \rightarrow 2\text{Cr}_2\text{O}_3\). (b) 0.256 g of \(\text{Cr}_2\text{O}_3\) is produced. (c) 0.081 g of \(\text{O}_2\) is required.

Step by step solution

01

Write the Unbalanced Equation

The reaction involves chromium (Cr) reacting with oxygen (O鈧�) to form chromium(III) oxide (Cr鈧侽鈧�). The unbalanced chemical equation is: \[\text{Cr (s) + O}_2\text{ (g)} \rightarrow \text{Cr}_2\text{O}_3\text{ (s)}\]
02

Balance the Equation

To balance the equation, ensure the same number of each type of atom on both sides. The balanced equation is: \[4\text{Cr (s) + 3O}_2\text{ (g)} \rightarrow 2\text{Cr}_2\text{O}_3\text{ (s)}\]
03

Calculate Moles of Chromium

First, calculate the moles of chromium. The atomic mass of chromium (Cr) is 51.996 g/mol. Moles of Cr = \(\frac{0.175 \text{ g}}{51.996 \text{ g/mol}} \approx 0.00337 \text{ mol}\)
04

Use Stoichiometry to Find Moles of Cr鈧侽鈧�

From the balanced equation, 4 moles of Cr produce 2 moles of Cr鈧侽鈧�. Therefore, \(0.00337 \text{ mol Cr} \times \frac{2 \text{ mol Cr}_2\text{O}_3}{4 \text{ mol Cr}} = 0.001685 \text{ mol Cr}_2\text{O}_3\).
05

Calculate Mass of Cr鈧侽鈧�

The molar mass of Cr鈧侽鈧� is calculated from its formula: \(2(51.996) + 3(16.00) = 151.992 \text{ g/mol}\). The mass of Cr鈧侽鈧� is therefore: \(0.001685 \text{ mol} \times 151.992 \text{ g/mol} = 0.256 \text{ g}\).
06

Find Mass of O鈧� Required

From the balanced equation, 4 moles of Cr require 3 moles of O鈧�. Calculate moles of O鈧� needed: \(0.00337 \text{ mol Cr} \times \frac{3 \text{ mol O}_2}{4 \text{ mol Cr}} = 0.0025275 \text{ mol O}_2\). Finally, calculate the mass of O鈧�: \(0.0025275 \text{ mol} \times 32.00 \text{ g/mol} = 0.081 \text{ g O}_2\).

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 91影视!

Key Concepts

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

Understanding Balanced Chemical Equations
A balanced chemical equation is essential for understanding the stoichiometry of a reaction. It ensures that the number of atoms of each element is the same on both the reactant and product sides of the equation. This balance is crucial because it follows the law of conservation of mass; matter is neither created nor destroyed in a chemical reaction.
  • When writing a chemical equation, begin with the unbalanced equation. For example, the reaction between chromium metal and oxygen is represented as: \[\text{Cr (s) + O}_2\text{ (g)} \rightarrow \text{Cr}_2\text{O}_3\text{ (s)}\]
  • To balance it, compare the number of atoms of each element on both sides. If they aren't equal, adjust coefficients accordingly. In this case: \[4\text{Cr (s) + 3O}_2\text{ (g)} \rightarrow 2\text{Cr}_2\text{O}_3\text{ (s)}\]
In the balanced form, the equation shows that 4 moles of chromium react with 3 moles of oxygen to produce 2 moles of chromium(III) oxide. This ratio is vital for calculations involving moles and mass.
Mole Calculations in Stoichiometry
The concept of moles is fundamental in stoichiometry as it allows chemists to count particles in a substance by weighing it. Knowing how to calculate moles is essential for determining other quantities in a reaction.
  • To find moles, use the formula: \[\text{Moles} = \frac{\text{Mass in grams}}{\text{Molar mass in g/mol}}\]
  • For chromium, with a mass of 0.175 g and a molar mass of 51.996 g/mol, the moles are calculated as: \[\frac{0.175 \text{ g}}{51.996 \text{ g/mol}} \approx 0.00337 \text{ mol}\]
This value is used in stoichiometric calculations to determine how much of a product is formed or how much of a reactant is needed. This step is crucial because it directly links mass to the chemical equation through the concept of moles.
Calculating Mass in Chemical Reactions
Mass calculations are an integral part of stoichiometry, where the mass of reactants and products must be determined from the given quantities.
  • Using a known amount of chromium, the mass of chromium(III) oxide formed can be found by first converting moles of Cr to moles of Cr鈧侽鈧� using the balanced chemical equation.
  • From the equation, 4 moles of Cr yields 2 moles of Cr鈧侽鈧�. Therefore, 0.00337 mol of Cr forms 0.001685 mol of Cr鈧侽鈧�.
  • Then, use the molar mass of Cr鈧侽鈧� (151.992 g/mol) to find the mass: \[0.001685 \text{ mol} \times 151.992 \text{ g/mol} = 0.256 \text{ g}\]
This calculation helps determine how much product is formed from a given amount of reactant, enabling predictions and adjustments in chemical processes.

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

Aqueous solutions of iron(II) chloride and sodium sulfide react to form iron(II) sulfide and sodium chloride. (a) Write the balanced equation for the reaction. (b) If you combine \(40 .\) g each of \(\mathrm{Na}_{2} \mathrm{S}\) and \(\mathrm{FeCl}_{2}\) what is the limiting reactant? (c) What mass of FeS is produced? (d) What mass of \(\mathrm{Na}_{2} \mathrm{S}\) or \(\mathrm{FeCl}_{2}\) remains after the reaction? (e) What mass of \(\mathrm{FeCl}_{2}\) is required to react completely with \(40 .\) g of \(\mathrm{Na}_{2} \mathrm{S}\) ?

Identify the ions that exist in each aqueous solution, and specify the concentration of each ion. (a) \(0.25 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) (b) \(0.123 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) (c) \(0.056 \mathrm{M} \mathrm{HNO}_{3}\)

You wish to determine the weight percent of copper in a copper-containing alloy. After dissolving a \(0.251-g\) sample of the alloy in acid, an excess of KI is added, and the \(\mathrm{Cu}^{2+}\) and \(\mathrm{I}^{-}\) ions undergo the reaction $$2 \mathrm{Cu}^{2+}(\mathrm{aq})+5 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{CuI}(\mathrm{s})+\mathrm{I}_{3}^{-}(\mathrm{aq})$$ The liberated \(\mathrm{I}_{3}^{-}\) is titrated with sodium thiosulfate according to the equation $$\mathrm{I}_{3}^{-}(\mathrm{aq})+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq}) \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq})$$ (a) Designate the oxidizing and reducing agents in the two reactions above. (b) If \(26.32 \mathrm{mL}\) of \(0.101 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) is required for titration to the equivalence point, what is the weight percent of Cu in the alloy?

A solution of hydrochloric acid has a volume of \(125 \mathrm{mL}\) and a pH of \(2.56 .\) What mass of \(\mathrm{NaHCO}_{3}\) must be added to completely consume the HCl?

An unknown compound has the formula \(\mathrm{C}_{x} \mathrm{H}_{y} \mathrm{O}_{z}\) You burn \(0.1523 \mathrm{g}\) of the compound and isolate \(0.3718 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(0.1522 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula of the compound? If the molar mass is \(72.1 \mathrm{g} / \mathrm{mol},\) what is the molecular formula?
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

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.