Question

Challenge An excess of acetic acid is added to 28 g of sodium bicarbonate at 25°C and 1 atm pressure. During the reaction, the gas cools to 20°C. What volume of carbon dioxide will be produced? The balanced equation for the reaction is shown below. $$NaHCO_{3}(\mathrm{aq})+CH_{3} \mathrm{COOH}(\mathrm{aq}) \rightarrow NaCH_{3} \mathrm{COO}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})+H_{2} O(\mathrm{l})$$

Short answer

The volume of carbon dioxide produced during the reaction under the given conditions is approximately 7.98 L.

Step-by-step solution

Determine the moles of sodium bicarbonate reacting

First, we need to identify the molar mass of sodium bicarbonate (NaHCO₃). The molar masses of Na, H, C, and O are 22.99 g/mol, 1.01 g/mol, 12.01 g/mol, and 16.00 g/mol, respectively. Therefore, the molar mass of NaHCO₃ is given by: Molar mass of NaHCO₃ = (1 × 22.99) + (1 × 1.01) + (1 × 12.01) + (3 × 16.00) = 84.01 g/mol Now, let's find the number of moles (n) of sodium bicarbonate using the given mass (28 g) and the molar mass: n = mass / molar mass n = 28 g / 84.01 g/mol = 0.333 mol

Determine the moles of carbon dioxide produced

Now that we have the moles of sodium bicarbonate reacting, we can use stoichiometry from the balanced chemical equation to determine the moles of carbon dioxide produced. The balanced equation shows that 1 mole of NaHCO₃ reacts to produce 1 mole of CO₂. So, the moles of carbon dioxide produced (nCO₂) will be equal to the moles of sodium bicarbonate: nCO₂ = n = 0.333 mol

Use the Ideal Gas Law to determine the volume of carbon dioxide

Now, we need to determine the volume of carbon dioxide produced at the given conditions (20°C and 1 atm). We will use the Ideal Gas Law for this calculation: PV = nRT where P is the pressure (in atm), V is the volume (in L), n is the number of moles, R is the ideal gas constant (0.0821 L atm/mol K), and T is the temperature (in K). First, convert the given temperature from Celsius to Kelvin: T = 20°C + 273.15 = 293.15 K Now, plug the values into the Ideal Gas Law equation: (1 atm) × V = (0.333 mol) × (0.0821 L atm/mol K) × (293.15 K)

Calculate the volume of carbon dioxide

Now, we just need to solve the Ideal Gas Law equation for the volume (V): V = (0.333 mol × 0.0821 L atm/mol K × 293.15 K) / 1 atm V ≈ 7.98 L So, the volume of carbon dioxide produced during the reaction under the given conditions is approximately 7.98 L.

Key concepts

Molar Mass Calculation

Understanding the molar mass of a substance is crucial for interpreting chemical reactions and performing calculations involving moles. Molar mass represents the weight of one mole of a substance, usually expressed in grams per mole (g/mol). It's calculated by adding together the atomic masses of each element in a compound.
For example, in the reaction involving sodium bicarbonate, we calculate its molar mass by summing the atomic weights of sodium (Na), hydrogen (H), carbon (C), and oxygen (O), each multiplied by the number of atoms of that element present in the formula unit of the substance. Once we have the molar mass, it allows us to convert between the mass of a substance and the number of moles, which sets the foundation for stoichiometric calculations.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation that relates the pressure (P), volume (V), temperature (T), and the number of moles (n) of a gas. The law is commonly stated as PV = nRT, where R is the ideal gas constant. For this equation, it's important to note that temperature must be in Kelvin, which can be found by adding 273.15 to the Celsius temperature.
In practical scenarios, like the reaction between acetic acid and sodium bicarbonate, the Ideal Gas Law allows us to calculate one of the variables if the others are known. Typically, we solve for the volume of a gas—as is the case in our example—making it an invaluable tool for chemists. Remember to keep the pressure in atmospheres and the volume in liters to match the units of the gas constant, which in our case is 0.0821 L atm/mol K.

Stoichiometric Calculations

Stoichiometry is the section of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It allows us to predict the amounts of products and reactants that are consumed and produced. When we know the stoichiometry of a reaction, we can use the coefficients from the balanced equation to determine the mole-to-mole relationships between the substances. For instance, the balanced equation of sodium bicarbonate reacting with acetic acid tells us that one mole of sodium bicarbonate will produce one mole of carbon dioxide.
Using stoichiometric calculations, once the number of moles of one substance is known, the amounts of other substances involved can be calculated. This is essential in many aspects of chemistry, including material design, biochemical pathways, and industrial synthesis of chemicals.