Problem 9 - Chemistry and Chemical Reactivity

Question

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?

Step-by-Step Solution

Answer

(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 1: Write the Unbalanced Equation

The reaction involves chromium (Cr) reacting with oxygen (O₂) to form chromium(III) oxide (Cr₂O₃). The unbalanced chemical equation is: \[\text{Cr (s) + O}_2\text{ (g)} \rightarrow \text{Cr}_2\text{O}_3\text{ (s)}\]

Step 2: Balance the Equation

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

Step 3: Calculate Moles of Chromium

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

Step 4: Use Stoichiometry to Find Moles of Cr₂O₃

From the balanced equation, 4 moles of Cr produce 2 moles of Cr₂O₃. Therefore, \(0.00337 \text{ mol Cr} \times \frac{2 \text{ mol Cr}_2\text{O}_3}{4 \text{ mol Cr}} = 0.001685 \text{ mol Cr}_2\text{O}_3\).

Step 5: Calculate Mass of Cr₂O₃

The molar mass of Cr₂O₃ is calculated from its formula: \(2(51.996) + 3(16.00) = 151.992 \text{ g/mol}\). The mass of Cr₂O₃ is therefore: \(0.001685 \text{ mol} \times 151.992 \text{ g/mol} = 0.256 \text{ g}\).

Step 6: Find Mass of O₂ Required

From the balanced equation, 4 moles of Cr require 3 moles of O₂. Calculate moles of O₂ 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 O₂: \(0.0025275 \text{ mol} \times 32.00 \text{ g/mol} = 0.081 \text{ g O}_2\).

Key Concepts

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 Cr₂O₃ using the balanced chemical equation.
• From the equation, 4 moles of Cr yields 2 moles of Cr₂O₃. Therefore, 0.00337 mol of Cr forms 0.001685 mol of Cr₂O₃.
• Then, use the molar mass of Cr₂O₃ (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.