Problem 135 - Chemistry and Chemical Reactivity

Question

Chromium is obtained by heating chromium(III) oxide with carbon. Calculate the mass percent of chromium in the oxide, and then use this value to calculate the quantity of \(\mathrm{Cr}_{2} \mathrm{O}_{3}\) required to produce \(850 \mathrm{kg}\) of chromium metal.

Step-by-Step Solution

Answer

Mass percent of chromium in Cr₂O₃ is 68.42%; about 1242.97 kg of Cr₂O₃ is needed.

Step 1: Determine the Molar Mass of Cr₂O₃

Chromium(III) oxide (Cr₂O₃) consists of 2 chromium (Cr) atoms and 3 oxygen (O) atoms. The atomic mass of Cr is approximately 52.00 amu, and O is approximately 16.00 amu. Calculate the molar mass:\[ M_{\text{Cr}_2\text{O}_3} = 2 \times 52.00 + 3 \times 16.00 = 104.00 + 48.00 = 152.00 \, \text{g/mol} \]

Step 2: Calculate the Mass of Chromium in Cr₂O₃

In the formula Cr₂O₃, there are 2 Cr atoms. Therefore, the mass contribution from chromium is:\[ \text{Mass of Cr} = 2 \times 52.00 = 104.00 \, \text{g/mol}\]

Step 3: Calculate Mass Percent of Chromium in Cr₂O₃

The mass percent of chromium in Cr₂O₃ is calculated by dividing the mass of chromium by the molar mass of Cr₂O₃, then multiplying by 100:\[ \text{Mass percent of Cr} = \left( \frac{104.00}{152.00} \right) \times 100 \approx 68.42\%\]

Step 4: Determine Mass of Cr₂O₃ Required to Produce 850 kg of Cr

Since 68.42% of Cr₂O₃ is chromium, use this information to find how much Cr₂O₃ is needed to obtain 850 kg of chromium:\[ \text{Mass of Cr}_2\text{O}_3 = \frac{850}{0.6842} \approx 1242.97 \, \text{kg}\]

Key Concepts

Molar Mass Calculation

When dealing with chemical compounds like chromium(III) oxide (Cr₂O₃), it is important to determine the molar mass. This concept enables us to convert between mass and number of moles, providing a foundation for various calculations in chemistry.

The formula for chromium(III) oxide shows that it consists of two chromium (Cr) atoms and three oxygen (O) atoms. Each chromium atom contributes about 52 amu and each oxygen atom contributes about 16 amu. The total molar mass of Cr₂O₃ is calculated by summing the contributions of all atoms involved:

• Chromium: 2 atoms × 52.00 amu = 104.00 amu
• Oxygen: 3 atoms × 16.00 amu = 48.00 amu

Putting it all together, the molar mass of chromium(III) oxide equals 152.00 g/mol.

Mass Percent Calculation

Mass percent is an expression of the concentration of an element in a compound. It represents the mass of a specific element divided by the total mass of the compound, multiplied by 100 to convert into a percentage. For determining the mass percent of chromiumin chromium(III) oxide, use the following steps.

First, calculate the mass of chromium present in the compound. Two chromium atoms contribute a total mass of 104.00 g/mol. Then, divide the mass of chromium by the molar mass of the entire compound, Cr₂O₃.

This gives:\[ \text{Mass percent of Cr} = \left( \frac{104.00}{152.00} \right) \times 100 \approx 68.42\%.\]This indicates that about 68.42% of the mass of chromium(III) oxide is due to chromium.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It is essentially a quantitative aspect of chemistry that is used to predict the amount of substances consumed and produced in a given reaction. With the knowledge of the mass percent of chromium in chromium(III) oxide, we can calculate how much of the oxide is needed to produce a specific amount of pure chromium.

Since we know that 68.42% of Cr₂O₃ is chromium, we can establish how much Cr₂O₃ is necessary to obtain 850 kg of chromium:\[\text{Mass of Cr}_2\text{O}_3 = \frac{850 \, \text{kg}}{0.6842} \approx 1242.97 \, \text{kg}.\]This indicates that approximately 1242.97 kg of chromium(III) oxide is needed to produce 850 kg of chromium. Stoichiometry helps us manage these precise calculations.

Chemical Reactions

Chemical reactions are at the heart of processes such as the production of chromium from chromium(III) oxide. These reactions involve the transformation of substances through the breaking and forming of chemical bonds. In this case, chromium(III) oxide is reduced using carbon, resulting in the formation of chromium metal and carbon dioxide.

This can be represented by the balanced chemical reaction:\[2 \text{Cr}_2\text{O}_3 + 3 \text{C} \rightarrow 4 \text{Cr} + 3 \text{CO}_2\]Understanding these reactions requires recognizing the stoichiometric ratios of reactants and products. The coefficients in the balanced equation revealthe molar relationships between different substances involved.

This reaction highlights the importance of balancing chemical equationsand comprehending molar relationships to predict the results of chemical processes accurately.By breaking these concepts into manageable steps, students can gain a better understanding of howelements from the periodic table interact in various chemical reactions, especially in industry-related processes.