Problem 61 - Chemistry The Science in Context

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 \(25.0 \mathrm{g}\) of \(\mathrm{NaHCO}_{3}\).

Step-by-Step Solution

Answer

Answer: The mass of CO2 produced from the decomposition of 25.0 g of NaHCO3 is 6.54 g.

Step 1: Write the balanced chemical equation for the decomposition reaction

To solve this, we first write the balanced chemical equation for the decomposition reaction of NaHCO3: NaHCO3(s) -> Na2CO3(s) + H2O(g) + CO2(g) On checking the atoms on both sides of the arrow, we find that there are 2 sodium atoms on the product side, and just 1 sodium atom on the reactant side. Since the rest of the elements are balanced, we can fix this by adding a 2 in front of NaHCO3, making the balanced equation: 2 NaHCO3(s) -> Na2CO3(s) + H2O(g) + CO2(g)

Step 2: Conversion of quantities

Before calculating the mass of CO2 produced, we need to know the relationship between the moles of NaHCO3 and CO2. For this, check the balanced equation above: 2 NaHCO3(s) -> Na2CO3(s) + H2O(g) + CO2(g) From the balanced equation, we can observe that 2 moles of NaHCO3 decompose to form 1 mole of CO2. This means that the mole ratio between NaHCO3 and CO2 is 2:1. Now, we will convert the given mass of NaHCO3 into moles using its molar mass: Molar mass of NaHCO3 = 23.0 (Na) + 1.01 (H) + 12.01 (C) + 3(16.00) (O) = 84.01 g/mol

Step 3: Calculate the mass of CO2

To calculate the mass of CO2 produced, first, convert the 25.0 g of NaHCO3 into moles: Moles of NaHCO3 = 25.0 g / 84.01 g/mol = 0.2975 mol Next, use the mole ratio (2:1) to calculate the moles of CO2 formed: Moles of CO2 = 0.2975 mol NaHCO3 × (1 mol CO2 / 2 mol NaHCO3) = 0.1487 mol CO2 Finally, convert the moles of CO2 into mass using its molar mass: Molar mass of CO2 = 12.01 (C) + 2(16.00) (O) = 44.01 g/mol Mass of CO2 = 0.1487 mol × 44.01 g/mol = 6.54 g So, the mass of CO2 produced from the decomposition of 25.0 g of NaHCO3 is 6.54 g.

Key Concepts

Understanding Balancing Chemical Equations

Balancing chemical equations is essential for understanding chemical reactions. In a balanced chemical equation, the number of each type of atom must be the same on both sides of the reaction equation. This demonstrates the law of conservation of mass, which states that matter cannot be created or destroyed.

During the decomposition of sodium bicarbonate (\(\text{NaHCO}_3\)), we begin by writing an unbalanced equation: \(\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)\). After observing the sodium atoms on both sides, it's clear we're missing a sodium atom in the reactant. Thus, placing a coefficient of 2 before \(\text{NaHCO}_3\) balances the equation:

• \(2 \text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)\)

Now, each atom has the same count on both sides: two sodium (Na), two carbon (C), two hydrogen (H), and six oxygen (O) atoms are balanced. This step ensures that the reaction complies with the conservation of mass.

The Mole Concept Explained

The mole concept is a fundamental tool in chemistry for converting between atoms/molecules and grams. One mole of any substance contains Avogadro's number (\(6.022 \times 10^{23}\)) of atoms or molecules. This allows chemists to work with these vast numbers in a manageable way.

In the decomposition of \(\text{NaHCO}_3\), we aim to find how much carbon dioxide gas (\(\text{CO}_2\)) is produced. By first converting the given mass of \(\text{NaHCO}_3\) (25.0 g) to moles, you use its molar mass: 84.01 g/mol. This calculation is represented by:

• Moles of \(\text{NaHCO}_3 = \frac{25.0 \text{ g}}{84.01 \text{ g/mol}} = 0.2975 \text{ mol}\)

Understanding moles helps bridge the gap between the microscopic world of atoms and the macroscopic world we observe, allowing precise chemical quantification.

Mastering Stoichiometry

Stoichiometry involves quantifying the reactants and products in a chemical reaction. It depends on a balanced chemical equation and the mole concept.

Once the balanced equation for the decomposition of \(\text{NaHCO}_3\) is established, stoichiometry clarifies the proportional relationships in the equation. For every 2 moles of \(\text{NaHCO}_3\) decomposed, 1 mole of \(\text{CO}_2\) is produced, found as:

• Moles of \(\text{CO}_2 = 0.2975 \text{ mol } \text{NaHCO}_3 \times \frac{1 \text{ mol } \text{CO}_2}{2 \text{ mol } \text{NaHCO}_3} = 0.1487 \text{ mol } \text{CO}_2\)

Finally, converting moles of \(\text{CO}_2\) to mass using its molar mass (44.01 g/mol) provides the result:

• Mass of \(\text{CO}_2 = 0.1487 \text{ mol} \times 44.01 \text{ g/mol} = 6.54 \text{ g}\)

Stoichiometry allows us to calculate how much of a substance is required or produced in a reaction, making it invaluable for both laboratory experiments and industrial processes.