Problem 104 - Chemistry and Chemical Reactivity

Question

An Alka-Seltzer tablet contains exactly \(100 . \mathrm{mg}\) of citric acid, \(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7},\) plus some sodium bicarbonate. What mass of sodium bicarbonate is required to consume \(100 .\) mg of citric acid by the following reaction? $$\begin{aligned}\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(\mathrm{aq})+3 \mathrm{NaHCO}_{3}(\mathrm{aq}) \rightarrow & \\\3 \mathrm{H}_{2} \mathrm{O}(\ell)+3 \mathrm{CO}_{2}(\mathrm{g}) &+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(\mathrm{aq})\end{aligned}$$

Step-by-Step Solution

Answer

131 mg of sodium bicarbonate is required.

Step 1: Understand the Chemical Reaction

The balanced chemical reaction given is: \( \mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7} + 3 \mathrm{NaHCO}_{3} \rightarrow 3 \mathrm{H}_{2} \mathrm{O} + 3 \mathrm{CO}_{2} + \mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7} \). This indicates that 1 mole of citric acid reacts with 3 moles of sodium bicarbonate.

Step 2: Calculate Molar Masses

The molar mass of citric acid, \( \mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7} \), is calculated as follows: C (12.01) \(*\) 6 + H (1.01) \(*\) 8 + O (16.00) \(*\) 7 = 192.12 \ \text{g/mol}. The molar mass of sodium bicarbonate, \( \mathrm{NaHCO}_{3} \), is: Na (22.99) + H (1.01) + C (12.01) + O (16.00) \(*\) 3 = 84.01 \ \text{g/mol}.

Step 3: Convert Milligrams to Grams

Convert the mass of citric acid from milligrams to grams: \(100 \, \text{mg} = 0.100 \, \text{g}\).

Step 4: Determine Moles of Citric Acid

Using the molar mass of citric acid, calculate the number of moles: \( \text{moles of citric acid} = \frac{0.100 \ \text{g}}{192.12 \ \text{g/mol}} \approx 0.00052 \text{ moles} \).

Step 5: Calculate Moles of Sodium Bicarbonate Required

The reaction stoichiometry shows 3 moles of \( \mathrm{NaHCO}_{3} \) are needed per mole of citric acid. Hence, \( \text{moles of NaHCO}_3 = 3 \times 0.00052 \approx 0.00156 \text{ moles} \).

Step 6: Convert Moles of Sodium Bicarbonate to Grams

Using the molar mass of \( \mathrm{NaHCO}_{3} \), convert moles to grams: \( \text{mass of NaHCO}_3 = 0.00156 \text{ moles} \times 84.01 \ \text{g/mol} \approx 0.131 \ \text{g} \) or 131 mg.

Key Concepts

Chemical Reaction

A chemical reaction involves the transformation of reactants into products. In this specific reaction, citric acid and sodium bicarbonate come together to form water, carbon dioxide, and trisodium citrate. It is crucial to write the balanced chemical equation to understand the exact relationship between the reactants and products. A balanced equation indicates that the number of atoms for each element is the same on both sides of the reaction. Here, 1 mole of citric acid \[ (\mathrm{H}_3\mathrm{C}_6\mathrm{H}_5\mathrm{O}_7) \] reacts with 3 moles of sodium bicarbonate \[ (\mathrm{NaHCO}_3) \] to produce water, carbon dioxide, and trisodium citrate \( (\mathrm{Na}_3\mathrm{C}_6\mathrm{H}_5\mathrm{O}_7) \). Understanding the chemical reaction sets the stage for further calculations.

Molar Mass

Molar mass is the mass of one mole of a substance and is often expressed in grams per mole (g/mol). It serves as a conversion factor between the mass of a substance and the amount of moles it contains. To find the molar mass, sum up the atomic masses of all the atoms in the chemical formula. For citric acid \((\mathrm{H}_3\mathrm{C}_6\mathrm{H}_5\mathrm{O}_7)\), the molar mass is calculated by adding:

• Carbon: 6 atoms, 12.01 g/mol each, total 72.06 g/mol
• Hydrogen: 8 atoms, 1.01 g/mol each, total 8.08 g/mol
• Oxygen: 7 atoms, 16.00 g/mol each, total 112.00 g/mol

Summing these gives 192.12 g/mol for citric acid. Similarly, for sodium bicarbonate \((\mathrm{NaHCO}_3)\), the molar mass is calculated by adding:

• Sodium: 22.99 g/mol
• Hydrogen: 1.01 g/mol
• Carbon: 12.01 g/mol
• Oxygen: 3 atoms, 16.00 g/mol each, total 48.00 g/mol

This sums up to 84.01 g/mol for sodium bicarbonate. Molar mass is a key element in determining how much of a reactant or product is needed or produced in a chemical reaction.

Conversion Calculations

Conversion calculations involve changing quantities from one unit to another to facilitate understanding of a chemical process. In this problem, we start with the known mass of citric acid in milligrams. First, convert the citric acid from milligrams to grams, which requires dividing by 1000, because \(1000 \, \text{mg} = 1 \, \text{g}\). Thus, that means \(100 \, \text{mg} = 0.100 \, \text{g}\).Next, use the molar mass of citric acid \((192.12 \, \text{g/mol})\) for conversion to moles, applying the formula:\[\text{Moles of citric acid} = \frac{\text{Mass in grams}}{\text{Molar mass}} = \frac{0.100 \text{g}}{192.12 \text{g/mol}} \approx 0.00052 \, \text{moles}\]This result allows us to determine the amount of sodium bicarbonate needed using the mole ratio from the balanced equation.

Stoichiometric Coefficients

Stoichiometric coefficients are the numbers located before compounds in a balanced chemical reaction. They tell us the proportional relationships between reactants and products. In the given reaction equation:\[\mathrm{H}_3\mathrm{C}_6\mathrm{H}_5\mathrm{O}_7 + 3 \mathrm{NaHCO}_3 \rightarrow 3 \mathrm{H}_2\mathrm{O} + 3 \mathrm{CO}_2 + \mathrm{Na}_3\mathrm{C}_6\mathrm{H}_5\mathrm{O}_7\]The coefficients indicate for every molar amount of citric acid, three times that amount of sodium bicarbonate is required. For our converted moles of citric acid \(0.00052\), you multiply by the stoichiometric coefficient for sodium bicarbonate:\[\text{Moles of } \mathrm{NaHCO}_3 = 3 \times 0.00052 \approx 0.00156 \, \text{moles}\]To find the mass of sodium bicarbonate, convert moles back to grams:\[\text{Mass of } \mathrm{NaHCO}_3 = 0.00156 \, \text{moles} \times 84.01 \, \text{g/mol} \approx 0.131 \, \text{g}\]or 131 mg. The stoichiometric coefficients are essential for ensuring that the reactants are used in the correct proportions needed for the reaction to take place efficiently.