Question

When sodium bicarbonate, \(\mathrm{NaHCO}_{3}(s),\) is heated, sodium carbonate is produced, with the evolution of water vapor and carbon dioxide gas. $$ 2 \mathrm{NaHCO}_{3}(s) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g) $$ What total volume of gas, measured at 29 " \(\mathrm{C}\) and 769 torr, is produced when \(1.00 \mathrm{~g}\) of \(\mathrm{NaHCO}_{3}(s)\) is completely converted to \(\mathrm{Na}_{2} \mathrm{CO}_{3}(s) ?\)

Short answer

The total volume of gas (water vapor and carbon dioxide) produced when 1.00 g of sodium bicarbonate is completely converted to sodium carbonate at 29°C and 769 torr is approximately 0.288 L.

Step-by-step solution

Calculate the number of moles of sodium bicarbonate

To find the number of moles, we will use the mass of sodium bicarbonate and its molar mass. The molar mass of \(\mathrm{NaHCO}_{3}\) is: M(Na) + M(H) + M(C) + 3 * M(O) = 22.99 + 1.01 + 12.01 + 3 * 16.00 = 84.01 g/mol Now, divide the mass of sodium bicarbonate by its molar mass: Number of moles of \(\mathrm{NaHCO}_{3}\) = \(\frac{1.00 \thinspace g}{84.01 \thinspace g/mol}\) = 0.0119 mol

Determine the number of moles of water vapor and carbon dioxide produced

From the balanced chemical equation, it is evident that for every 2 moles of \(\mathrm{NaHCO}_{3}\), 1 mole of \(\mathrm{H}_{2}\mathrm{O}\) and 1 mole of \(\mathrm{CO}_{2}\) are produced. Therefore, the number of moles of \(\mathrm{H}_{2}\mathrm{O}\) and \(\mathrm{CO}_{2}\) will be half of the moles of \(\mathrm{NaHCO}_{3}\) in this case. Number of moles of \(\mathrm{H}_{2}\mathrm{O}\) = \(\frac{1}{2} ×\) 0.0119 mol = 0.00595 mol Number of moles of \(\mathrm{CO}_{2}\) = \(\frac{1}{2} ×\) 0.0119 mol = 0.00595 mol

Use the ideal gas law to find the total volume of the combined gases

The ideal gas law is given by: PV = nRT where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. We are given the temperature, 29 °C, which we need to convert to Kelvin: T = 29 °C + 273.15 = 302.15 K Pressure is given in torr (769 torr), so we'll need to convert it to atm to use the appropriate gas constant (R = 0.0821 atm L/mol K): P = \(\frac{769 \thinspace torr}{760 \thinspace torr/atm}\) = 1.012 atm The total number of moles of gas (both water vapor and carbon dioxide) is 0.00595 + 0.00595 = 0.0119 mol. Now we can substitute all the given information into the ideal gas law formula: (1.012 atm)(V) = (0.0119 mol)(0.0821 atm L/mol K)(302.15 K) Now, solve for V: V = \(\frac{(0.0119 \thinspace mol)(0.0821 \thinspace atm \thinspace L/mol \thinspace K)(302.15 \thinspace K)}{1.012 \thinspace atm}\) = 0.288 L So, the total volume of gas produced when \(1.00 \thinspace g\) of \(\mathrm{NaHCO}_{3}\) is completely converted to \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) at 29 °C and 769 torr is approximately 0.288 L.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a critical formula in chemistry that helps us understand the behavior of gases under various conditions. It is often represented as \(PV = nRT\). Let's break down each part of this formula:

• \(P\) stands for pressure. In the case of our exercise, the pressure is initially given in torr, a common unit of measurement. To use the standard gas constant, it's necessary to convert torr into atmospheres (atm).
• \(V\) is the volume of the gas, which is what we are solving for in the problem.
• \(n\) represents the number of moles of gas. In our exercise, we found this by using the initial mass of \(\mathrm{NaHCO}_{3}\) and the balanced chemical equation.
• \(R\) is the ideal gas constant. For the units atm, L, mol, and K, \(R\)'s value is 0.0821 \(\mathrm{atm \cdot L \cdot mol^{-1} \cdot K^{-1}}\).
• \(T\) is the temperature in Kelvin, which ensures consistent unit usage.

Understanding how each component interacts allows us to solve for any unknown when given sufficient details about the other variables. This law assumes gases behave ideally, which usually holds true under conditions of low pressure and high temperature.

Chemical Reactions

Chemical reactions are processes where substances, called reactants, are transformed into different substances, called products. In our exercise, we are dealing with the thermal decomposition of sodium bicarbonate, \(\mathrm{NaHCO}_{3}\).
Here are some key points about this reaction:

• The reaction is balanced, meaning the number of atoms for each element is the same on both sides of the equation. This follows the law of conservation of mass.
• For every 2 moles of \(\mathrm{NaHCO}_{3}\), 1 mole of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\), 1 mole of \(\mathrm{H}_{2}\mathrm{O}(g)\), and 1 mole of \(\mathrm{CO}_{2}(g)\) are produced. Understanding stoichiometry helps us make these connections.
• The reaction involves phase changes: sodium bicarbonate changes from a solid to other forms including gases.

By analyzing the stoichiometry of chemical reactions, we can predict quantities like the amount of gas produced, which is essential for calculations in various chemical processes.

Mole Calculations

Mole calculations are a foundational skill in chemistry, helping to connect the mass of substances to the number of particles, such as atoms or molecules. The mole is a standard unit in chemistry for counting a very large number of small entities. Here's how we use moles in our problem:

• First, we calculate the molar mass of sodium bicarbonate (\(\mathrm{NaHCO}_3\)). This involves summing the atomic masses of its constituent elements.
• With the molar mass known, we can determine the number of moles in a given mass of \(\mathrm{NaHCO}_3\) using the formula: Number of moles = \(\frac{\text{mass}}{\text{molar mass}}\).
• This calculation reveals how many moles of each product are created, based on the balanced chemical equation.

Using moles ensures that our chemical calculations are both accurate and precise, allowing us to scale reactions up or down based on need. It's a bridge that connects theoretical chemistry to practical lab work.