Problem 79 - Chemistry The Central Science In Si Units

Question

The fizz produced when an Alka-Seltzer tablet is dissolved in water is due to the reaction between sodium bicarbonate \(\left(\mathrm{NaHCO}_{3}\right)\) and citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)\) $$ \begin{aligned} 3 \mathrm{NaHCO}_{3}(a q)+\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) & \longrightarrow \\ & 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) \end{aligned} $$ In a certain experiment \(1.00 \mathrm{~g}\) of sodium bicarbonate and \(1.00 \mathrm{~g}\) of citric acid are allowed to react.(a) Which is the limiting reactant? (b) How many grams of carbon dioxide form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed?

Step-by-Step Solution

Answer

(a) Sodium bicarbonate is the limiting reactant. (b) 0.524 g of CO₂ is produced. (c) 0.236 g of citric acid remains.

Step 1: Calculate moles of sodium bicarbonate

First, we need to calculate the number of moles of sodium bicarbonate \(\left( \text{NaHCO}_3 \right)\). Its molar mass is approximately \(84.01 \, \text{g/mol}\). Using the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \] we calculate: \[ \text{moles of NaHCO}_3 = \frac{1.00 \, \text{g}}{84.01 \, \text{g/mol}} \approx 0.0119 \, \text{mol} \].

Step 2: Calculate moles of citric acid

Next, calculate the moles of citric acid \(\left( \text{H}_3\text{C}_6\text{H}_5\text{O}_7 \right)\), which has a molar mass of approximately \(192.12 \, \text{g/mol}\). Using the same formula: \[ \text{moles of H}_3\text{C}_6\text{H}_5\text{O}_7 = \frac{1.00 \, \text{g}}{192.12 \, \text{g/mol}} \approx 0.0052 \, \text{mol} \].

Step 3: Determine the limiting reactant

The balanced equation is \[3 \text{NaHCO}_3 + \text{H}_3\text{C}_6\text{H}_5\text{O}_7 \to 3 \text{CO}_2 + 3 \text{H}_2\text{O} + \text{Na}_3\text{C}_6\text{H}_5\text{O}_7\]. The mole ratio of \(\text{NaHCO}_3\) to \(\text{H}_3\text{C}_6\text{H}_5\text{O}_7\) is 3:1. \(0.0119\) moles of \(\text{NaHCO}_3\) would require \(\frac{0.0119}{3} \approx 0.0040\) moles of \(\text{H}_3\text{C}_6\text{H}_5\text{O}_7\), but we have more than enough \(\text{H}_3\text{C}_6\text{H}_5\text{O}_7\). Therefore, \(\text{NaHCO}_3\) is the limiting reactant.

Step 4: Calculate grams of carbon dioxide formed

Based on the stoichiometry of the reaction, 3 moles of \(\text{NaHCO}_3\) produce 3 moles of \(\text{CO}_2\). Thus 0.0119 moles of \(\text{NaHCO}_3\) will produce 0.0119 moles of \(\text{CO}_2\). The molar mass of \(\text{CO}_2\) is \(44.01 \, \text{g/mol}\), so the mass of \(\text{CO}_2\) is \[ \text{Mass of CO}_2 = 0.0119 \, \text{mol} \times 44.01 \, \text{g/mol} \approx 0.524 \, \text{g} \].

Step 5: Calculate the excess reactant remaining

Initially, there were 0.0052 moles of \(\text{H}_3\text{C}_6\text{H}_5\text{O}_7\). As \(\text{NaHCO}_3\) was the limiting reactant, \(0.00397\) moles of \(\text{H}_3\text{C}_6\text{H}_5\text{O}_7\) were used (calculated as \(\frac{0.0119}{3}\)). Therefore, \(0.0052 - 0.00397 \approx 0.00123\) moles of \(\text{H}_3\text{C}_6\text{H}_5\text{O}_7\) remain. The remaining mass is \[ 0.00123 \, \text{mol} \times 192.12 \, \text{g/mol} \approx 0.236 \, \text{g} \].

Key Concepts

Molar Mass

Molar mass is an essential concept in chemistry that represents the mass of one mole of a given substance. It is expressed in units of grams per mole (g/mol). To calculate the molar mass of a compound, you add up the molar masses of all the atoms present, which can be found on the periodic table. For instance, sodium bicarbonate (NaHCO₃) has a molar mass of about 84.01 g/mol, and citric acid (H₃C₆H₅O₇) has a molar mass of approximately 192.12 g/mol. These values are pivotal when determining how much of each substance will react. Whenever dealing with chemical reactions, molar mass allows you to convert grams into moles, making stoichiometric calculations possible.

Stoichiometry

Stoichiometry is the part of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It relies heavily on the balanced chemical equation, which shows the ratio of molecules that participate in the reaction. For example, in the reaction between sodium bicarbonate and citric acid, the balanced equation is:\[ 3 ext{NaHCO}_3 + ext{H}_3 ext{C}_6 ext{H}_5 ext{O}_7 o 3 ext{CO}_2 + 3 ext{H}_2 ext{O} + ext{Na}_3 ext{C}_6 ext{H}_5 ext{O}_7 \].This tells us that three moles of sodium bicarbonate react with one mole of citric acid to produce three moles of carbon dioxide, among other products. Understanding stoichiometry enables us to determine amounts of reactants consumed and products formed, using mole ratios derived from the balanced equation.

Chemical Reactions

Chemical reactions occur when substances interact to form new products. In our exercise, the fizzing reaction of an Alka-Seltzer tablet in water involves sodium bicarbonate and citric acid reacting to form carbon dioxide, water, and a sodium citrate compound. Chemical equations symbolize these transformations, signifying reactants and products along with their molar relationships. Each compound involved reacts in a specific ratio, as depicted by the equation. Identifying the balancing of such equations is crucial, as it ensures that the number of atoms for each element is the same on both sides of the equation, maintaining the conservation of mass principle.

Excess Reactant

An excess reactant is a substance that remains after a chemical reaction has completed. It is not completely used up because there was more than enough of it to react with the limiting reactant. In chemical processes, identifying the excess reactant is advantageous, as it can help optimize industrial processes or reduce waste. In the given reaction, citric acid is the excess reactant, as the sodium bicarbonate is fully consumed. Once you know the limiting reactant, it's straightforward to calculate how much of the other reactants remain unreacted. This often involves subtracting the moles that reacted from the initial moles, translating that into grams if necessary.