Based on the percentages of components in Alka-Seltzer and the balanced equation above, determine the limiting reactant assuming 1 gram of Alka-Seltzer.

Question

Group of answer choices. 3 NaHCO₃ (aq) + C₆H₈O₇ (aq) → 3 CO₂ (g) + Na₃C₆H₅O₇ (aq) + 3 H₂O (l)

Answer 1

To determine the limiting reactant in the reaction provided, we first need to identify the percentages of the components in Alka-Seltzer. Alka-Seltzer typically contains sodium bicarbonate (NaHCO₃) and citric acid (C₆H₈O₇) as its main active ingredients.

Step 1: Determine the masses of the reactants

Assuming 1 gram of Alka-Seltzer, we need to know the approximate percentage of each reactant. For the sake of this example, let's use hypothetical percentages:

• Sodium bicarbonate (NaHCO₃): 60%
• Citric acid (C₆H₈O₇): 40%

So, for 1 gram of Alka-Seltzer:

• Mass of NaHCO₃ = 0.60 g
• Mass of C₆H₈O₇ = 0.40 g

Step 2: Convert grams to moles

Next, we convert these masses to moles using their molar masses.

• Molar mass of NaHCO₃ ≈ 84.01 g/mol
• Molar mass of C₆H₈O₇ ≈ 192.13 g/mol

For NaHCO₃: \[ \text{Moles of NaHCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{0.60 , \text{g}}{84.01 , \text{g/mol}} \approx 0.00714 , \text{mol} \]

For C₆H₈O₇: \[ \text{Moles of C}_6H_8O_7 = \frac{0.40 , \text{g}}{192.13 , \text{g/mol}} \approx 0.00208 , \text{mol} \]

Step 3: Use the stoichiometry of the balanced equation

The balanced reaction is: \[ 3 \text{NaHCO}_3 + \text{C}_6H_8O_7 \rightarrow 3 \text{CO}_2 + \text{Na}_3\text{C}_6\text{H}_5\text{O}_7 + 3 \text{H}_2\text{O} \]

From the balanced equation:

• 3 moles of NaHCO₃ react with 1 mole of C₆H₈O₇.

Step 4: Determine the molar ratio

Using the moles calculated:

• The stoichiometric ratio from the equation indicates that for every 3 moles of NaHCO₃, 1 mole of C₆H₈O₇ is needed.

From our number of moles:

• We have 0.00714 moles of NaHCO₃ and 0.00208 moles of C₆H₈O₇.

Now, calculate how many moles of NaHCO₃ are needed to react with the available C₆H₈O₇: \[ \text{Moles of NaHCO}_3 \text{ needed} = 3 \times \text{moles of C}_6H_8O_7 = 3 \times 0.00208 \approx 0.00624 , \text{mol} \]

Step 5: Compare available and required moles

• Available NaHCO₃: 0.00714 mol
• Required NaHCO₃ for the available C₆H₈O₇: 0.00624 mol

Since we have more available NaHCO₃ than needed (0.00714 mol vs 0.00624 mol), C₆H₈O₇ will limit the reaction.

Conclusion

C₆H₈O₇ (citric acid) is the limiting reactant in this scenario.