Problem 33 - Chemistry: An Atoms-Focused Approach

Question

When \(\mathrm{NaHCO}_{3}\) is heated above \(270^{\circ} \mathrm{C},\) it decomposes to \(\mathrm{Na}_{2} \mathrm{CO}_{3}(s), \mathrm{H}_{2} \mathrm{O}(g),\) and \(\mathrm{CO}_{2}(g)\) a. Write a balanced chemical equation for the decomposition reaction. b. Calculate the mass of \(\mathrm{CO}_{2}\) produced from the decomposition of \(25.0 \mathrm{g} \mathrm{NaHCO}_{3}.\)

Step-by-Step Solution

Answer

Answer: Approximately 6.55 g of CO2 is produced from the decomposition of 25.0 g of NaHCO3.

Step 1: Balancing the chemical equation

The reaction is given by the following (unbalanced) equation: NaHCO3(s) --> Na2CO3(s) + H2O(g) + CO2(g) In order to balance the chemical equation, we search for the appropriate coefficients that will make the number of atoms on both sides of the equation equal. We do this as follows: 2 NaHCO3(s) --> Na2CO3(s) + H2O(g) + CO2(g) So, the balanced chemical equation is: 2 NaHCO3(s) --> Na2CO3(s) + H2O(g) + CO2(g)

Step 2: Calculating the molar mass of NaHCO3 and CO2

Before we can determine the mass of CO2 produced, we need to calculate the molar mass of NaHCO3 and CO2. Using the periodic table, we find that: Molar mass of NaHCO3 = 22.99 g/mol (Na) + 1.01 g/mol (H) + 12.01 g/mol (C) + 48.00 g/mol (O3) = 84.01 g/mol Molar mass of CO2 = 12.01 g/mol (C) + 32.00 g/mol (O2) = 44.01 g/mol

Step 3: Using stoichiometry to find the mass of CO2 produced

From the balanced equation, we can see that 2 moles of NaHCO3 decompose to produce 1 mole of CO2. We will now use this stoichiometric relationship to find the mass of CO2 produced from the decomposition of 25.0 g of NaHCO3. First, we calculate the moles of NaHCO3 in 25.0 g: moles of NaHCO3 = (mass of NaHCO3) / (molar mass of NaHCO3) = (25.0 g) / (84.01 g/mol) = 0.2977 mol Next, we find the moles of CO2 produced using the stoichiometric relationship: moles of CO2 = (0.2977 mol NaHCO3) * (1 mol CO2 / 2 mol NaHCO3) = 0.1489 mol CO2 Finally, we calculate the mass of CO2 produced: mass of CO2 = (moles of CO2) * (molar mass of CO2) = (0.1489 mol) * (44.01 g/mol) = 6.5547 g So, the mass of CO2 produced from the decomposition of 25.0 g of NaHCO3 is approximately 6.55 g.

Key Concepts

Stoichiometry

Understanding stoichiometry is like learning the recipe of a chemical reaction. It involves figuring out how reactants (ingredients) and products (results) relate to each other in a chemical equation. In our example, when heating sodium bicarbonate, or NaHCO₃, it turns into sodium carbonate (Na₂CO₃), water (H₂O), and carbon dioxide (CO₂). The chemical reaction isn't meaningful until it's balanced. This simply means having the same number of each type of atom on both sides of the reaction.

For example, the balanced reaction for our decomposition is:

• 2 NaHCO₃ (s) ––> Na₂CO₃ (s) + H₂O (g) + CO₂ (g)

From this, you might see two sodium bicarbonate molecules producing one molecule of sodium carbonate, one molecule of water, and one molecule of carbon dioxide.

This concept, Stoichiometry, helps us predict the amount of product when we know the amount of reactants, allowing chemists to budget materials effectively and understand the full outcomes of reactions.

Molar Mass Calculation

Molar mass calculation is essential to convert between the mass of a substance and the number of moles. Molar mass tells you how much one mole of a chemical compound weighs. It's like the chemical version of checking your grocery receipt for how much each item weighs! For NaHCO₃, molar mass calculation involves understanding the individual weights of sodium (Na), hydrogen (H), carbon (C), and oxygen (O).

• Sodium (Na) = 22.99 g/mol
• Hydrogen (H) = 1.01 g/mol
• Carbon (C) = 12.01 g/mol
• Oxygen (O₃) = 48.00 g/mol

The total molar mass of NaHCO₃, adding all these together, is 84.01 g/mol.

For CO₂:

• Carbon (C) = 12.01 g/mol
• Oxygen (O₂) = 32.00 g/mol

Making 44.01 g/mol for CO₂.

By using molar masses, we can relate grams of a substance to the moles. This is crucial for stoichiometry, as reactions often give you moles needed, but real-world examples might measure out grams.

Balancing Chemical Equations

Balancing chemical equations is at the core of understanding reactions like the decomposition of NaHCO₃. Each side of the equation must have the same number of each atom to satisfy the Law of Conservation of Mass. This law insists that mass can't be created or destroyed. If we have two sodium atoms on the reactant side, we need two on the product side too.

To balance an equation:

• Count the atoms of each element in the reactants and products.
• Adjust coefficients (the numbers in front of molecules, not subscripts) to achieve the same number of each type of atom on both sides.
• Make sure your final coefficients are the smallest set of whole numbers.

In our decomposition of sodium bicarbonate example:

• The unbalanced equation initially: NaHCO₃(s) ––> Na₂CO₃(s) + H₂O(g) + CO₂(g)
• Becomes balanced as: 2 NaHCO₃(s) ––> Na₂CO₃(s) + H₂O(g) + CO₂(g)

Balancing ensures the reaction complies with physical laws and represents the actual chemical change. It's like ensuring you're using the right amounts of each ingredient when baking a cake so it turns out perfectly!