Question

How many milliliters of carbon dioxide gas at STP are produced from the decomposition of \(5.00 \mathrm{~g}\) of iron(III) carbonate? $$\mathrm{Fe}_{2}\left(\mathrm{CO}_{3}\right)_{3}(s) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3\mathrm{CO}_{2}(g)$$

Short answer

1,152.13 mL of \( \text{CO}_2 \) are produced.

Step-by-step solution

Write the Balanced Chemical Equation

The decomposition reaction of iron(III) carbonate is given as \( \mathrm{Fe}_{2}\left(\mathrm{CO}_{3}\right)_{3}(s) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3\mathrm{CO}_{2}(g) \). This equation shows that 1 mole of iron(III) carbonate produces 3 moles of carbon dioxide gas.

Calculate Molar Mass of Iron(III) Carbonate

To find the molar mass of \( \text{Fe}_2(\text{CO}_3)_3 \), calculate: 2 moles of Fe (\(2 \times 55.85 \ \text{g/mol}\)), 3 moles of C (\(3 \times 12.01 \ \text{g/mol}\)), and 9 moles of O (\(9 \times 16.00 \ \text{g/mol}\)). Thus, the molar mass = \( 2(55.85) + 3(12.01) + 9(16.00) = 291.72 \ \text{g/mol} \).

Calculate Moles of Iron(III) Carbonate

Use the mass of iron(III) carbonate to calculate moles: \( \frac{5.00 \ \text{g}}{291.72 \ \text{g/mol}} = 0.01714 \ \text{mol} \).

Calculate Moles of Carbon Dioxide Produced

From the stoichiometry of the balanced equation, 1 mole of \( \text{Fe}_2(\text{CO}_3)_3 \) produces 3 moles of \( \text{CO}_2 \). Therefore, \( 0.01714 \ \text{mol} \times 3 = 0.05142 \ \text{mol} \) of \( \text{CO}_2 \) is produced.

Convert Moles of Carbon Dioxide to Volume at STP

At STP, 1 mole of any gas occupies 22.4 liters, or 22,400 milliliters. Thus, \( 0.05142 \ \text{mol} \times 22,400 \ \text{mL/mol} = 1,152.13 \ \text{mL} \) of \( \text{CO}_2 \).

Key concepts

Molar Mass Calculation

Molar mass calculation is a crucial step in solving chemical problems, as it allows us to convert between the mass of a substance and the amount in moles. To calculate the molar mass, we sum the masses of all atoms present in a formula. For iron(III) carbonate, represented by the formula \( \mathrm{Fe}_2(\mathrm{CO}_3)_3 \), the calculation involves adding the masses of iron, carbon, and oxygen atoms included in each molecule.

• Iron (Fe): There are 2 atoms, with a mass of about 55.85 g/mol each, so the total mass from iron is \( 2 \times 55.85 \ \text{g/mol} \).
• Carbon (C): Present in 3 molecules of carbonate, each containing one carbon atom, giving a total of \( 3 \times 12.01 \ \text{g/mol} \).
• Oxygen (O): Each carbonate group contains three O atoms, totaling 9 atoms. Thus, the oxygen contribution is \( 9 \times 16.00 \ \text{g/mol} \).

Adding these, the molar mass of \( \mathrm{Fe}_2(\mathrm{CO}_3)_3 \) is found to be 291.72 g/mol. Accurate molar mass determination is vital in stoichiometry, especially when determining how much reactant is needed or product will be formed.

Gas Volume at STP

At standard temperature and pressure (STP), the behavior of gases is predictable using basic principles. One key rule is that 1 mole of any ideal gas occupies a volume of 22.4 liters or 22,400 milliliters at STP. This standardization makes it easier to relate the moles of gas in reactions to measurable volumes.When it comes to our reaction, we calculated the moles of carbon dioxide produced from the decomposition of iron(III) carbonate. Knowing that the decomposition produced 0.05142 moles of \( \text{CO}_2 \), we used the volume per mole at STP to find that:\[ 0.05142 \, \mathrm{mol} \times 22,400 \, \mathrm{mL/mol} = 1,152.13 \, \mathrm{mL} \]This result shows how much gas will fill under standard conditions, valuable information for experiments and industrial applications. Gas volume calculations are invaluable for chemists since they facilitate the prediction and measurement of gaseous by-products in a reaction.

Chemical Reaction Balancing

Balancing chemical reactions is an essential skill in chemistry as it ensures the conservation of mass and proper stoichiometric relationships between reactants and products. In the decomposition of iron(III) carbonate to form iron(III) oxide and carbon dioxide, the balanced equation is:\[\mathrm{Fe}_{2}(\mathrm{CO}_{3})_{3}(s) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s) + 3\mathrm{CO}_{2}(g)\]In this equation, each side contains the same number of each type of atom, demonstrating the conservation of mass principle:

• 2 Iron (Fe)
• 3 Carbon (C)
• 9 Oxygen (O)

Balancing allows us to understand the stoichiometry of the reaction, showing that 1 mole of \( \text{Fe}_2(\text{CO}_3)_3 \) yields 3 moles of \( \text{CO}_2 \). This direct relationship aids in predicting the amounts of products formed in a reaction. Mastering chemical reaction balancing underpins many other chemical calculations, emphasizing its significance in learning chemistry.