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) \(4 \text{Cr} + 3 \text{O}_2 \rightarrow 2 \text{Cr}_2\text{O}_3\). (b) 0.256 g \(\text{Cr}_2\text{O}_3\). (c) 0.081 g \(\text{O}_2\).

Step 1: Write the Unbalanced Chemical Equation

Start by writing the skeletal equation for the reaction of chromium with oxygen. The reaction involves chromium metal reacting with oxygen gas to form chromium(III) oxide, \(\text{Cr}_2\text{O}_3\). The unbalanced equation is: \[\text{4 Cr} + \text{3 O}_2 \rightarrow \text{2 Cr}_2\text{O}_3\]

Step 2: Balance the Chemical Equation

Next, balance the chemical equation by making sure that the number of atoms for each element is equal on both sides of the reaction. Balance chromium first: there are 2 chromium atoms on the product side, so use 4 chromium atoms on the reactant side. Then, balance oxygen by using 3 oxygen molecules (\(\text{O}_2\)) on the reactant side. The balanced equation is: \[\text{4 Cr} + \text{3 O}_2 \rightarrow \text{2 Cr}_2\text{O}_3\]

Step 3: Calculate Moles of Chromium

To find the mass of \(\text{Cr}_2\text{O}_3\) produced, begin by calculating the number of moles of chromium. The molar mass of chromium (\(\text{Cr}\)) is \(51.9961\) grams/mole. Use the equation for moles: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\).\[\text{Moles of Cr} = \frac{0.175 \text{ g}}{51.9961 \text{ g/mol}} \approx 0.00337 \text{ mol}\]

Step 4: Relate Moles of Chromium to Moles of \(\text{Cr}_2\text{O}_3\)

From the balanced equation, 4 moles of chromium yield 2 moles of \(\text{Cr}_2\text{O}_3\). Therefore, for \(0.00337\) moles of chromium, calculate moles of \(\text{Cr}_2\text{O}_3\):\[\text{Moles of } \text{Cr}_2\text{O}_3 = \frac{0.00337 \text{ mol Cr}}{4} \times 2 \approx 0.001685 \text{ mol}\]

Step 5: Calculate Mass of \(\text{Cr}_2\text{O}_3\) Produced

Use the moles of \(\text{Cr}_2\text{O}_3\) to find the mass. The molar mass of \(\text{Cr}_2\text{O}_3\) is \(151.99\) grams/mole. Calculate the mass using: \(\text{mass} = \text{moles} \times \text{molar mass}\).\[\text{Mass of } \text{Cr}_2\text{O}_3 = 0.001685 \text{ mol} \times 151.99 \text{ g/mol} \approx 0.256 \text{ g}\]

Step 6: Calculate Moles of Oxygen Required

From the balanced equation, 3 moles of \(\text{O}_2\) react with 4 moles of chromium, which is \(0.00337\) moles. Calculate moles of \(\text{O}_2\) needed.\[\text{Moles of } \text{O}_2 = \frac{0.00337 \text{ mol Cr}}{4} \times 3 \approx 0.002528 \text{ mol}\]

Step 7: Calculate Mass of Oxygen Required

Use the moles of oxygen to calculate the mass needed. The molar mass of \(\text{O}_2\) is \(32.00\) grams/mole.\[\text{Mass of } \text{O}_2 = 0.002528 \text{ mol} \times 32.00 \text{ g/mol} \approx 0.081 \text{ g}\]

Key Concepts

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into new substances, called products. These transformations involve rearranging atoms to form new chemical bonds. In our exercise, the reaction involves the transformation of chromium metal (\( \text{Cr} \)) and oxygen gas (\( \text{O}_2 \)) into chromium(III) oxide (\( \text{Cr}_2\text{O}_3 \)). This reaction can be described using a chemical equation: \[\text{4 Cr} + \text{3 O}_2 \rightarrow \text{2 Cr}_2\text{O}_3\]This representation shows the reactants on the left and the products on the right. The chemical equation is a succinct way to express the chemical reaction.

• Reactants: Starting materials in a reaction (e.g., chromium and oxygen).
• Products: Substances formed by the reaction (e.g., chromium(III) oxide).

Understanding chemical reactions is fundamental as they are central to chemistry and are involved in countless processes, from cooking to industrial manufacturing.

Molecular Weight Calculation

To understand the weight of chemicals involved in reactions, we calculate molecular weights. The molecular weight (or molar mass) of a compound is the sum of the atomic weights of all the atoms in its formula. In this exercise, for example, we consider the molar mass of chromium (\( \text{Cr} \)), which is \( 51.9961 \) grams/mole.For more complex compounds, such as chromium(III) oxide (\( \text{Cr}_2\text{O}_3 \)), we use the atomic weights of its constituent elements:

• Chromium (\( \text{Cr} \)): \( 51.9961 \) grams/mole.
• Oxygen (\( \text{O} \)): \( 16.00 \) grams/mole.

The molar mass of \( \text{Cr}_2\text{O}_3 \) is calculated as:\[2 \times 51.9961 \text{ g/mol} + 3 \times 16.00 \text{ g/mol} = 151.99 \text{ g/mol}\]Calculating the molecular weight allows us to determine the amount of each substance involved in a chemical reaction.

Balancing Chemical Equations

Balancing chemical equations is crucial for depicting a reaction accurately. It involves making sure there are equal numbers of each type of atom on both sides of the equation. This ensures that matter is conserved according to the Law of Conservation of Mass.In our reaction involving chromium and oxygen, the balancing process is as follows:

• Start with the initial unbalanced equation: \( \text{4 Cr} + \text{3 O}_2 \rightarrow \text{2 Cr}_2\text{O}_3 \)
• Count the atoms of each element on both sides.
• Adjust coefficients, not subscripts, to have the same number of each type of atom on both sides.

Here, we equalize the number of chromium and oxygen atoms:

• Use 4 chromium atoms to ensure the same number on both sides.
• Use 3 oxygen molecules (\( \text{O}_2 \)) to balance the oxygen atoms.

Once balanced, the equation correctly represents how atoms are rearranged during the process. This skill is foundational for solving any stoichiometry problem in chemistry.