Question

Pure bismuth can be produced by the reaction of bismuth oxide with carbon at high temperatures. $$2 \mathrm{Bi}_{2} \mathrm{O}_{3}+3 \mathrm{C} \rightarrow 4 \mathrm{Bi}+3 \mathrm{CO}_{2}$$ How many moles of \(\mathrm{Bi}_{2} \mathrm{O}_{3}\) reacted to produce 12.6 \(\mathrm{mol}\) of \(\mathrm{CO}_{2} ?(\text {Chapter } 11)\)

Short answer

8.4 moles of Bi₂O₃ reacted to produce 12.6 moles of CO₂.

Step-by-step solution

Write down the balanced chemical equation

The balanced chemical equation is given as: \[2 \mathrm{Bi}_{2} \mathrm{O}_{3}+3 \mathrm{C} \rightarrow 4 \mathrm{Bi}+3 \mathrm{CO}_{2}\]

Calculate the mole-to-mole ratio

From the balanced chemical equation, we can see that for every 3 moles of CO₂ produced, 2 moles of Bi₂O₃ react. The mole-to-mole ratio between Bi₂O₃ and CO₂ is: \[\frac{2 \ \text{moles of Bi}_{2}\text{O}_{3}}{3 \ \text{moles of CO}_{2}}\]

Calculate the required moles of Bi₂O₃

Now we need to find the number of moles of Bi₂O₃ required to produce 12.6 moles of CO₂. We can do this by multiplying the mole-to-mole ratio by the given moles of CO₂. \[n_{\mathrm{Bi}_{2}\mathrm{O}_{3}} = \frac{2 \ \text{moles of Bi}_{2}\text{O}_{3}}{3 \ \text{moles of CO}_{2}} \times 12.6 \ \text{moles of CO}_{2}\]

Perform the calculation

Calculating the moles of Bi₂O₃: \[n_{\mathrm{Bi}_{2}\mathrm{O}_{3}} = \frac{2}{3} \times 12.6 = 8.4\] So, 8.4 moles of Bi₂O₃ reacted to produce 12.6 moles of CO₂.

Key concepts

Stoichiometry

Stoichiometry is a section of chemistry that focuses on the quantifiable relationships between the reactants and products in a chemical reaction. Using stoichiometry, we can predict how much of each substance is needed or produced, based solely on the chemical equation at hand.

Consider it like recipe ratios for cooking, but instead of dealing with cups and teaspoons, chemists deal with moles, which is a measure of the number of particles in a substance. The mole-to-mole ratio, extracted from the coefficients of a balanced chemical equation, is the core of stoichiometry and governs the proportions in which chemicals react.

Balanced Chemical Equation

A balanced chemical equation is vital because it shows the correct proportions of reactants and products, ensuring the law of conservation of mass is followed - matter is neither created nor destroyed during a chemical reaction. In your exercise, the equation \[2 \mathrm{Bi}_{2} \mathrm{O}_{3}+3 \mathrm{C} \rightarrow 4 \mathrm{Bi}+3 \mathrm{CO}_{2}\] is balanced, meaning that the number of atoms for each element is equal on both sides of the equation.

This balance is what allows us to derive mole-to-mole ratios that can be used to calculate how much of one chemical is needed to react completely with another. Without a balanced equation, these calculations would not be accurate, leading to incorrect conclusions about the reaction.

Molecular Conversion Calculations

Molecular conversion calculations are the step-by-step processes we use to convert between the moles of different substances in a reaction. By using the mole-to-mole ratios obtained from the balanced chemical equation, we can determine the amount of any reactant required to produce a desired quantity of product, or vice versa.

For example, in the provided solution, the calculation \[n_{\mathrm{Bi}_{2}\mathrm{O}_{3}} = \frac{2}{3} \times 12.6 = 8.4\] used the determined mole-to-mole ratio to find the moles of \(\mathrm{Bi}_{2} \mathrm{O}_{3}\) needed to produce 12.6 moles of \(\mathrm{CO}_{2}\). This illustrates how stoichiometry allows chemists to scale reactions up or down and predict outcomes in the laboratory or industrial settings.