Problem 26 - Chemistry and Chemical Reactivity

Question

At higher temperatures, NaHCO is converted quantitatively to \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) $$ 2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ Heating a \(1.7184-\mathrm{g}\) sample of impure NaHCO \(_{3}\) gives \(0.196 \mathrm{g}\) of \(\mathrm{CO}_{2} .\) What was the mass percent of \(\mathrm{NaHCO}_{3}\) in the original 1.7184 -g sample?

Step-by-Step Solution

Answer

The mass percent of \( \mathrm{NaHCO}_{3} \) is approximately 43.47\%.

Step 1: Calculate Moles of \( \mathrm{CO}_{2} \) Produced

The molar mass of \( \mathrm{CO}_{2} \) is approximately 44.01 g/mol. Calculate the moles of \( \mathrm{CO}_{2} \) using the mass given:\[\text{Moles of } \mathrm{CO}_{2} = \frac{0.196 \text{ g}}{44.01 \text{ g/mol}} \approx 0.00445 \text{ mol}\]

Step 2: Relate \( \mathrm{CO}_{2} \) to \( \mathrm{NaHCO}_{3} \)

The balanced chemical equation shows that 2 moles of \( \mathrm{NaHCO}_{3} \) produce 1 mole of \( \mathrm{CO}_{2} \). Therefore, the moles of \( \mathrm{NaHCO}_{3} \) that reacted are:\[\text{Moles of } \mathrm{NaHCO}_{3} = 2 \times 0.00445 \text{ mol} = 0.0089 \text{ mol}\]

Step 3: Find Mass of \( \mathrm{NaHCO}_{3} \) That Reacted

The molar mass of \( \mathrm{NaHCO}_{3} \) is approximately 84.01 g/mol. Calculate the mass of \( \mathrm{NaHCO}_{3} \) that reacted:\[\text{Mass of } \mathrm{NaHCO}_{3} = 0.0089 \text{ mol} \times 84.01 \text{ g/mol} \approx 0.747 \text{ g}\]

Step 4: Determine Mass Percent of \( \mathrm{NaHCO}_{3} \)

Calculate the mass percent of \( \mathrm{NaHCO}_{3} \) in the original sample using the mass of \( \mathrm{NaHCO}_{3} \) that reacted:\[\text{Mass percent of } \mathrm{NaHCO}_{3} = \left(\frac{0.747 \text{ g}}{1.7184 \text{ g}}\right) \times 100\% \approx 43.47\%\]

Key Concepts

chemical reactions

In chemistry, a chemical reaction is a process where substances, known as reactants, transform into different substances, called products. This transformation involves breaking of bonds in reactants and formation of new bonds in products. For instance, when heating sodium bicarbonate (\(\mathrm{NaHCO}_3\)), it decomposes into sodium carbonate (\(\mathrm{Na}_2\mathrm{CO}_3\)), carbon dioxide (\(\mathrm{CO}_2\)), and water (\(\mathrm{H}_2\mathrm{O}\)). During this process, energy is absorbed to break the bonds, and released when new ones are formed.

Chemical reactions can be influenced by various factors such as temperature, concentration of reactants, and even the presence of catalysts. Increasing the temperature generally increases the rate of a chemical reaction, allowing atoms to collide more energetically, leading to successful bond formation and a faster reaction.

It's important to identify the type of chemical reaction such as decomposition, synthesis, single-replacement, or double-replacement, to predict the products formed. Knowing these details enables chemists to manipulate conditions to achieve desired reactions efficiently.

moles and molar mass

The mole is a unit that measures the amount of substance. It is one of the foundational concepts in chemistry, essential for quantifying entities at the molecular and atomic scale. One mole of any substance contains Avogadro's number of particles, approximately \(6.022 \times 10^{23}\) entities, whether they be atoms, molecules, or ions.

Molar mass is the mass of one mole of a substance. It is usually expressed in grams per mole (g/mol). For example, the molar mass of carbon dioxide (\(\mathrm{CO}_2\)) is approximately 44.01 g/mol, calculated by adding the atomic masses of carbon (around 12.01 g/mol) and oxygen (about 16.00 g/mol each for two oxygen atoms).

• To find the number of moles from the mass, use the formula \( \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \).
• Conversely, to find mass from moles, rearrange the formula to \( \text{Mass} = \text{Moles} \times \text{Molar Mass} \).

Understanding moles and molar mass allows chemists to determine reactant quantities needed or product amounts formed in chemical reactions, crucial for both theoretical calculations and practical laboratory work.

chemical equations

Chemical equations represent chemical reactions using symbols and formulas. They provide a detailed accounting of the reactants and products involved. For balancing chemical equations, it is essential because mass and atoms are conserved in a reaction. This means the same number of each type of atom must appear on both sides of the equation.

In the given reaction of sodium bicarbonate (\(\mathrm{NaHCO}_3\)), the balanced chemical equation is:\(2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s}) + \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)This equation indicates that two molecules of \(\mathrm{NaHCO}_3\) decompose to form one molecule of \(\mathrm{Na}_2\mathrm{CO}_3\), one molecule of \(\mathrm{CO}_2\), and one molecule of \(\mathrm{H}_2\mathrm{O}\).

• Coefficients in a chemical equation tell us the proportion of molecules involved in the reaction.
• Reactants are listed on the left, while products are listed on the right.
• It's crucial to correctly balance an equation to respect the law of conservation of mass.

By correctly interpreting and using chemical equations, chemists can predict how substances will react, determine yield, and ensure safety in processes.