Question

\(\mathrm{NaHCO}_{3}\) decomposes when exposed to heat: \(2 \mathrm{NaHCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell) \Delta H=91.5 \mathrm{~kJ}\) What mass of \(\mathrm{NaHCO}_{3}\) is decomposed by \(256 \mathrm{~kJ}\) ?

Short answer

Approximately 469.6 g of \(\mathrm{NaHCO}_{3}\) decomposes by 256 kJ.

Step-by-step solution

Identify the given data and what is required

The chemical reaction given is the decomposition of \(\mathrm{NaHCO}_{3}\). We need to find the mass of \(\mathrm{NaHCO}_{3}\) that decomposes when \(256 \mathrm{~kJ}\) of energy is supplied. \(\Delta H\) for the decomposition of two moles of \(\mathrm{NaHCO}_{3}\) is \(91.5 \mathrm{~kJ}\).

Calculate moles of \(\mathrm{NaHCO}_{3}\) decomposed by 256 kJ

With \(\Delta H = 91.5 \mathrm{~kJ}\) for 2 moles of \(\mathrm{NaHCO}_{3}\), find how many moles relate to \(256 \mathrm{~kJ}\) using:\[\text{Moles of } \mathrm{NaHCO}_{3} = \frac{256 \mathrm{~kJ}}{91.5 \mathrm{~kJ}/2} = \frac{256 \times 2}{91.5} \approx 5.59 \text{ moles}\]

Calculate the molar mass of \(\mathrm{NaHCO}_{3}\)

The molar mass of each element in \(\mathrm{NaHCO}_{3}\) is:- Na: \(23.0 \mathrm{~g/mol}\)- H: \(1.0 \mathrm{~g/mol}\)- C: \(12.0 \mathrm{~g/mol}\)- O: \(16.0 \mathrm{~g/mol} \times 3 = 48.0 \mathrm{~g/mol}\).Thus, the molar mass of \(\mathrm{NaHCO}_{3}\) is \(23.0 + 1.0 + 12.0 + 48.0 = 84.0 \mathrm{~g/mol}\).

Calculate the mass of \(\mathrm{NaHCO}_{3}\)

Using the number of moles calculated in Step 2 and the molar mass from Step 3, find the mass:\[\text{Mass of } \mathrm{NaHCO}_{3} = 5.59 \text{ moles} \times 84.0 \mathrm{~g/mol} \approx 469.56 \mathrm{~g}\]

Key concepts

Enthalpy Change

Enthalpy change is a crucial concept in understanding chemical reactions. It refers to the amount of energy consumed or released when a chemical reaction occurs. In the context of the decomposition of sodium bicarbonate \(\mathrm{NaHCO}_{3}\), we have an enthalpy change \(\Delta H\) of 91.5 kJ for the decomposition of two moles of \(\mathrm{NaHCO}_{3}\).

This means that 91.5 kJ of energy is absorbed in the process. Knowing the enthalpy change is essential, as it allows us to calculate how much energy is needed to decompose a specific amount of reactant.

• Enthalpy change helps measure a system's heat change at constant pressure.
• A positive \(\Delta H\) indicates that the process is endothermic, meaning heat is absorbed.

In this particular reaction, understanding \(\Delta H\) facilitates the calculation of the mass of \(\mathrm{NaHCO}_{3}\) that reacts with 256 kJ of energy, providing a direct link between energy changes and chemical processes.

Chemical Decomposition

Chemical decomposition involves breaking down a compound into simpler substances. In the case of \(\mathrm{NaHCO}_{3}\), it decomposes into sodium carbonate \(\mathrm{Na}_{2}\mathrm{CO}_{3}\), carbon dioxide \(\mathrm{CO}_{2}\), and water \(\mathrm{H}_{2}\mathrm{O}\) when heated. This process can be represented by the chemical equation: \[2 \mathrm{NaHCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{Na}_{2}\mathrm{CO}_{3}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}\mathrm{O}(\ell)\]

Decomposition is a common chemical reaction and is mostly endothermic, requiring heat to proceed. Therefore, understanding this concept helps in controlling chemical processes, enabling predictions about how a compound will break down.

• Decomposition reactions often require an input of energy (heat, light, etc.) to initiate.
• The breakdown of substances is significant in fields like chemistry and environmental science.

In this exercise, it's useful to understand the pieces a compound decomposes into and how this ties into other stoichiometric calculations, such as the mass of reactants needed for a reaction.

Molar Mass Calculation

Molar mass is the mass of one mole of a given substance and is expressed in grams per mole \(\mathrm{g/mol}\). Calculating molar mass is essential because it translates moles into grams, linking the microscopic world to the macroscopic one.

For \(\mathrm{NaHCO}_{3}\), the calculation involves adding up the atomic masses of its constituent elements:

• Sodium (Na): 23.0 \(\mathrm{g/mol}\)
• Hydrogen (H): 1.0 \(\mathrm{g/mol}\)
• Carbon (C): 12.0 \(\mathrm{g/mol}\)
• Oxygen (O): 16.0 \(\mathrm{g/mol}\) x 3 = 48.0 \(\mathrm{g/mol}\)

Combining these gives a molar mass of 84.0 \(\mathrm{g/mol}\) for \(\mathrm{NaHCO}_{3}\). Understanding molar mass is invaluable for:

• Scaling chemical reactions appropriately for desired yields.
• Converting between moles and grams in various calculations.

In our specific case, knowing the molar mass allows us to find the mass of \(\mathrm{NaHCO}_{3}\) needed when a specific amount of energy is provided, completing the full circle of our stoichiometric analysis.