Problem 62 - Chemistry Matter and Change

Question

Antacid Fizz When an antacid tablet dissolves in water, the fizz is due to a reaction between sodium hydrogen carbonate \(\left(\mathrm{NaHCO}_{3}\right),\) also called sodium bicarbonate, and citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)\) \begin{equation} 3 \mathrm{NaHCO}_{3}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(\mathrm{aq}) \rightarrow \end{equation} \begin{equation} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad3 \mathrm{CO}_{2}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(1)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(\mathrm{aq})\end{equation} How many moles of \(\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\) can be produced if one tablet containing 0.0119 \(\mathrm{mol}\) of \(\mathrm{NaHCO}_{3}\) is dissolved?

Step-by-Step Solution

Answer

0.00397 moles of \(\mathrm{Na}_{3} \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{O}_{7}\) are produced.

Step 1: Understand the Balanced Chemical Equation

The balanced chemical equation for the reaction is \[ 3 \mathrm{NaHCO}_{3}(\mathrm{aq}) + \mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5}\mathrm{O}_{7}(\mathrm{aq}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{g}) + 3 \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{Na}_{3}\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{O}_{7}(\mathrm{aq}) \]. This equation indicates that 3 moles of \(\mathrm{NaHCO}_{3}\) react with 1 mole of \(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5}\mathrm{O}_{7}\) to form 1 mole of \(\mathrm{Na}_{3} \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{O}_{7}\).

Step 2: Determine the Limiting Reactant

Since only the amount of \(\mathrm{NaHCO}_{3}\) is given, we assume no more \(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5}\mathrm{O}_{7}\) than necessary is present. Therefore, \(\mathrm{NaHCO}_{3}\) is our limiting reactant.

Step 3: Calculate Moles of Product Formed

From the balanced equation, 3 moles of \(\mathrm{NaHCO}_{3}\) produce 1 mole of \(\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5}\mathrm{O}_{7}\). We have 0.0119 moles of \(\mathrm{NaHCO}_{3}\). Using the ratio from the equation, the moles of \(\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5}\mathrm{O}_{7}\) produced will be \(\frac{0.0119}{3}\). Calculate this to determine the product moles.

Step 4: Perform the Calculation

Perform the division: \(\frac{0.0119}{3} = 0.00397\). Therefore, 0.00397 moles of \(\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5}\mathrm{O}_{7}\) will be produced.

Key Concepts

Stoichiometry

Stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in a chemical reaction.
In simpler terms, it helps you understand how much of one chemical you need to react with another to create a product. Think of it as the recipe of chemistry where each molecule and atom must come together in specific proportions.

• The balanced chemical equation acts as a "recipe" to guide these relationships.
• The coefficients in the chemical equations give the ratios of molecules involved.

In our exercise, the equation shows that 3 moles of sodium bicarbonate (NaHCO_3) react with 1 mole of citric acid (H_3C_6H_5O_7) to form 1 mole of trisodium citrate (Na_3C_6H_5O_7) alongside water and carbon dioxide. Understanding these proportions is crucial for determining how much product can be made from given reactants.

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is totally consumed when the chemical reaction is complete. This reactant limits the amount of products that can be formed.
It's much like running a factory; if you run out of one part, you can't make any more of the product, even if you have plenty of all other parts.

• Identify which reactant runs out first—this is the limiting reactant.
• Use the limiting reactant to calculate the exact amount of each product formed.

In our exercise, sodium bicarbonate is the limiting reactant because we only have a specific amount available (0.0119 moles). Thus, it dictates how much trisodium citrate is produced.

Balanced Chemical Equation

A balanced chemical equation provides a visual representation of a chemical reaction using chemical formulas. It ensures conservation of mass, meaning that the number of atoms of each element is the same on both sides of the equation.
Balancing chemical equations is essential for stoichiometry calculations.

• It shows the exact number of reactant molecules needed to form a certain number of product molecules.
• Maintains the law of conservation of mass and energy.

For the antacid fizz reaction, the balanced equation: \[ 3\mathrm{NaHCO}_3(\mathrm{aq}) + \mathrm{H}_3\mathrm{C}_6\mathrm{H}_5\mathrm{O}_7(\mathrm{aq}) \rightarrow 3 \mathrm{CO}_2(\mathrm{g}) + 3 \mathrm{H}_2\mathrm{O}(l) + \mathrm{Na}_3\mathrm{C}_6\mathrm{H}_5\mathrm{O}_7(\mathrm{aq}) \] guides our stoichiometric calculations.

Mole Calculations

Mole calculations are a fundamental aspect of chemistry that allow scientists to determine the exact amount of a substance required or produced in a reaction. The mole is a unit that measures the amount of a chemical substance.
It's akin to a "chemist's counting unit," similar to a dozen, but much larger. One mole equals Avogadro's number of particles: approximately 6.022 x 10²³.

• Determine how many moles of a substance are involved using stoichiometry.
• Apply mole ratios from the balanced equation to calculate products and reactants.

In the exercise, we use mole calculations to find that 0.0119  moles of NaHCO_3 leads to 0.00397 moles of Na_3C_6H_5O_7 being formed. This is achieved by dividing the moles of limiting reactant by the ratio obtained from the balanced equation, highlighting a key technique in solving chemical problem exercises.