Question

World production of chromite \(\left(\mathrm{FeCr}_{2} \mathrm{O}_{4}\right),\) the main ore of chromium, was \(1.97 \times 10^{7}\) metric tons in 2006 . To isolate chromium, a mixture of chromite and sodium carbonate is heated in air to form sodium chromate, iron(III) oxide, and carbon dioxide. The sodium chromate is dissolved in water, and this solution is acidified with sulfuric acid to produce the less soluble sodium dichromate. The sodium dichromate is filtered out and reduced with carbon to produce chromium(III) oxide, sodium carbonate, and carbon monoxide. The chromium(III) oxide is then reduced to chromium with aluminum metal. (a) Write balanced equations for each step. (b) What mass of chromium (in \(\mathrm{kg}\) ) could be prepared from the 2006 world production of chromite?

Short answer

The mass of chromium produced from world production of chromite in 2006 is 9.152 million kg.

Step-by-step solution

Balance the reaction for forming sodium chromate

The initial reaction for producing sodium chromate involves heating chromite \(\text{FeCr}_2 \text{O}_4\) with sodium carbonate \( \text{Na}_2\text{CO}_3 \) in the presence of oxygen (air). The balanced equation is: \[ 4 \text{FeCr}_2 \text{O}_4 + 8 \text{Na}_2 \text{CO}_3 + 7 \text{O}_2 \rightarrow 8 \text{Na}_2 \text{CrO}_4 + 2 \text{Fe}_2 \text{O}_3 + 8 \text{CO}_2 \]

Acidification to form sodium dichromate

The sodium chromate \( \text{Na}_2 \text{CrO}_4 \) is then dissolved in water and acidified with sulfuric acid \( \text{H}_2 \text{SO}_4 \) to form sodium dichromate \( \text{Na}_2 \text{Cr}_2 \text{O}_7 \). The balanced equation for this step is: \[ 2 \text{Na}_2 \text{CrO}_4 + \text{H}_2 \text{SO}_4 \rightarrow \text{Na}_2 \text{Cr}_2 \text{O}_7 + \text{Na}_2 \text{SO}_4 + \text{H}_2 \text{O} \]

Reduction of sodium dichromate

Sodium dichromate \( \text{Na}_2 \text{Cr}_2 \text{O}_7 \) is reduced by carbon \( \text{C} \) to produce chromium(III) oxide \( \text{Cr}_2 \text{O}_3 \), sodium carbonate \( \text{Na}_2 \text{CO}_3 \), and carbon monoxide \( \text{CO} \). The balanced equation is: \[ \text{Na}_2 \text{Cr}_2 \text{O}_7 + 2 \text{C} \rightarrow \text{Cr}_2 \text{O}_3 + \text{Na}_2 \text{CO}_3 + 2 \text{CO} \]

Reduction of chromium(III) oxide

Chromium(III) oxide \( \text{Cr}_2 \text{O}_3 \) is then reduced using aluminum \( \text{Al} \) to produce chromium \( \text{Cr} \) and aluminum oxide \( \text{Al}_2 \text{O}_3 \). The balanced equation for this reduction is: \[ \text{Cr}_2 \text{O}_3 + 2 \text{Al} \rightarrow 2 \text{Cr} + \text{Al}_2 \text{O}_3 \]

Calculate the mass of chromium produced

First, calculate the moles of chromite in the world production. Given the molar mass of \(\text{FeCr}_2 \text{O}_4\), \[ \text{FeCr}_2 \text{O}_4 = 55.85 + 2(51.996) + 4(16) = 223.84 \text{ g/mol} \] World production = \(1.97 \times 10^7 \text{ tons} = 1.97 \times 10^{10} \text{ kg} \). Find moles of chromite: \[ \text{moles of chromite} = \frac{1.97 \times 10^{10} \text{ kg}}{223.84 \text{ g/mol}} = 8.8 \times 10^7 \text{ mol} \] Each mole of \( \text{FeCr}_2 \text{O}_4 \) produces 2 moles of \( \text{Cr} \), and the molar mass of Cr is 52 g/mol. Thus, the mass of Cr produced: \[ \text{mass of Cr} = 8.8 \times 10^7 \text{ mol} \times 2 \times 52 \text{ g/mol} = 9.152 \times 10^9 \text{ g} \] Convert grams to kilograms: \[ 9.152 \times 10^6 \text{ kg} \]

Key concepts

Chromium Extraction

Chromium extraction involves several crucial steps that turn chromite into pure chromium through a series of carefully controlled chemical reactions. All of these steps highlight the meticulous process required to obtain usable chromium from its ore.

• First, the chromite ore \( \text{FeCr}_2 \text{O}_4 \) is mixed with sodium carbonate \( \text{Na}_2 \text{CO}_3 \) and heated in air, resulting in the formation of sodium chromate \( \text{Na}_2 \text{CrO}_4 \), iron(III) oxide \( \text{Fe}_2 \text{O}_3 \), and carbon dioxide \( \text{CO}_2 \).
• Second, the sodium chromate is dissolved in water and acidified with sulfuric acid, which produces less soluble sodium dichromate \( \text{Na}_2 \text{Cr}_2 \text{O}_7 \).
• Then, sodium dichromate is reduced using carbon to get chromium(III) oxide \( \text{Cr}_2 \text{O}_3 \).
• Finally, Chromium(III) oxide is reduced using aluminum metal to produce pure chromium metal \( \text{Cr} \).

Each of these steps requires precision for successful extraction, reflecting the importance of understanding chemical processes for industrial applications.

Balanced Chemical Equations

Balanced chemical equations are essential to accurately represent the chemical reactions involved in chromium extraction. They show the quantity of reactants and products, guaranteeing the law of conservation of mass.

• For the formation of sodium chromate:**
  The equation is balanced as: \[ 4 \text{FeCr}_2 \text{O}_4 + 8 \text{Na}_2 \text{CO}_3 + 7 \text{O}_2 \rightarrow 8 \text{Na}_2 \text{CrO}_4 + 2 \text{Fe}_2 \text{O}_3 + 8 \text{CO}_2 \]
• Acidifying sodium chromate**:
  \[ 2 \text{Na}_2 \text{CrO}_4 + \text{H}_2 \text{SO}_4 \rightarrow \text{Na}_2 \text{Cr}_2 \text{O}_7 + \text{Na}_2 \text{SO}_4 + \text{H}_2 \text{O} \]
• Reduction of sodium dichromate**:
  \[ \text{Na}_2 \text{Cr}_2 \text{O}_7 + 2 \text{C} \rightarrow \text{Cr}_2 \text{O}_3 + \text{Na}_2 \text{CO}_3 + 2 \text{CO} \]
• Reduction of chromium(III) oxide**:
  \[ \text{Cr}_2 \text{O}_3 + 2 \text{Al} \rightarrow 2 \text{Cr} + \text{Al}_2 \text{O}_3 \]

These steps not only ensure the transformation is chemically feasible but also help in quantifying the input and output substances accurately.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. In the context of chromium extraction, it helps us find the theoretical yield of chromium from a given amount of chromite.
**Step-by-Step Calculation:**

• First, we determine the molar mass of chromite \( \text{FeCr}_2 \text{O}_4 \). Summing up the atomic masses yields 223.84 g/mol.
• Then, using the world production amount given (1.97 x 10^^7 metric tons or 1.97 x 10^^10 kg), we calculate the moles of chromite:
  \[ \text{moles of chromite} = \frac{1.97 \times 10^{10} \text{ kg}}{223.84 \text{ g/mol}} = 8.8 \times 10^7 \text{ mol} \]
• Each mole of chromite produces 2 moles of chromium. Thus, the moles of chromium:
  \[ 8.8 \times 10^7 \text{ mol} \times 2 = 1.76 \times 10^8 \text{ mol} \]
• Finally, we convert moles of chromium into mass using its molar mass (52 g/mol):
  \[ \text{mass of Cr} = 1.76 \times 10^8 \text{ mol} \times 52 \text{ g/mol} = 9.152 \times 10^9 \text{ g} \approx 9.15 \times 10^6 \text{ kg} \]

This calculated mass represents the potential chromium production from the 2006 chromite supply, emphasizing the power of stoichiometric methods.

Industrial Chemistry

Industrial chemistry translates laboratory chemical processes into large-scale manufacturing, dealing with the practical aspects of chemical production. **Key Insights into Industrial Chromium Extraction:**

• **Efficiency and Scale:**
  Industrial processes for extracting chromium have to be scalable and efficient. Every step needs optimization to ensure maximum yield and minimal waste. This includes finely tuning conditions like temperature and pressure.
• **Safety:**
  Chemicals used, such as sulfuric acid and aluminum, must be handled carefully. Industrial facilities incorporate extensive safety measures to protect workers and the environment.
• **Economics:**
  The cost-effectiveness of the extraction process is critical. It considers the availability and price of raw materials (chromite and reagents), energy consumption, and labor costs.
• **Environmental Impact:**
  Industrial operations also focus on reducing harmful emissions (like CO2) and properly managing waste products (like Fe2O3 and Na2SO4). Regulatory compliance is mandatory to avoid environmental penalties.

Understanding these aspects of industrial chemistry is crucial for aspiring chemical engineers and chemists who aim to join this field. It bridges the gap between theoretical knowledge and real-world applications.