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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�: \(0.0025275 \text{ mol} \times 32.00 \text{ g/mol} = 0.081 \text{ g O}_2\).

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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.

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Most popular questions from this chapter

What volume of \(0.750 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\), in milliliters, is required to react completely with 1.00 L of \(2.25 \mathrm{M} \mathrm{NaCl}\) solution? The balanced equation is \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{NaCl}(\mathrm{aq}) \rightarrow \mathrm{PbCl}_{2}(\mathrm{s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})\)

A sample of limestone and other soil materials was heated, and the limestone decomposed to give calcium oxide and carbon dioxide. $$\mathrm{CaCO}_{3}(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})$$ A \(1.506-\mathrm{g}\) sample of limestone-containing material gave \(0.558 \mathrm{g}\) of \(\mathrm{CO}_{2}\), in addition to \(\mathrm{CaO}\), after being heated at a high temperature. What was the mass percent of \(\mathrm{CaCO}_{3}\) in the original sample?

ATOM ECONOMY: Ethylene oxide, \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O},\) is an important industrial chemical [as it is the starting place to make such important chemicals as ethylene glycol (antifreeze) and various polymers 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 g of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is used to synthesize \(762 \mathrm{g}\) of the product by the catalytic reaction?

The metals industry was a major source of air pollution years ago. One common process involved "roasting" metal sulfides in the air: $$2 \mathrm{PbS}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{PbO}(\mathrm{s})+2 \mathrm{SO}_{2}(\mathrm{g})$$ If 2.50 mol of \(\mathrm{PbS}\) is heated in air, what amount of \(\mathrm{O}_{2}\) is required for complete reaction? What amounts of \(\mathrm{PbO}\) and \(\mathrm{SO}_{2}\) are expected?

The reaction of iron(III) oxide with aluminum to give molten iron is known as the thermite reaction (page \(172)\). $$\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+2 \mathrm{Al}(\mathrm{s}) \rightarrow 2 \mathrm{Fe}(\ell)+\mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})$$ What amount of \(\mathrm{Al}\), in moles, is needed for complete reaction with 3.0 mol of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) ? What mass of \(\mathrm{Fe},\) in grams, can be produced?
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