Question

Based on the percentages of components in Alka-Seltzer and the balanced equation above, determine the limiting reactant assuming 1 gram of Alka-Seltzer.
Group of answer choices. 3 NaHCO₃ (aq) + C₆H₈O₇ (aq) → 3 CO₂ (g) + Na₃C₆H₅O₇ (aq) + 3 H₂O (l) Do not hypothesis

Answer 1

To determine the limiting reactant based on the given equation and assuming we have 1 gram of Alka-Seltzer, we first need to know the composition of Alka-Seltzer and then calculate how much of each reactant is present in the 1 gram sample.

Alka-Seltzer generally contains sodium bicarbonate (NaHCO₃) and citric acid (C₆H₈O₇). Although the exact percentages and weights can vary, let's assume a commonly cited formulation. For this explanation:

- Alka-Seltzer contains approximately 1 g of active ingredients typically made of:
- **Sodium Bicarbonate (NaHCO₃):** ~ 1.5 g
- **Citric Acid (C₆H₈O₇):** ~ 1 g

Using the balanced equation provided:

\[ 3 \text{NaHCO}_3 (aq) + \text{C}_6\text{H}_8\text{O}_7 (aq) \rightarrow 3 \text{CO}_2 (g) + \text{Na}_3\text{C}_6\text{H}_5\text{O}_7 (aq) + 3 \text{H}_2\text{O} (l) \]

### Step 1: Molar Mass Calculation
- Molar mass of NaHCO₃ = 22.99 (Na) + 1.01 (H) + 12.01 (C) + 3 * 16.00 (O) = 84.01 g/mol
- Molar mass of C₆H₈O₇ = 6 * 12.01 (C) + 8 * 1.01 (H) + 7 * 16.00 (O) = 192.13 g/mol

### Step 2: Determine Moles in 1g of Each Component
Suppose you determine the distribution and you find that in your 1 gram sample (for simplification):
- Let's say Alka-Seltzer contains about 0.5 g NaHCO₃ and 0.4 g C₆H₈O₇.

Calculating the number of moles:
- Moles of NaHCO₃ = \( \frac{0.5 \text{ g}}{84.01 \text{ g/mol}} \approx 0.00595 \text{ mol} \)
- Moles of C₆H₈O₇ = \( \frac{0.4 \text{ g}}{192.13 \text{ g/mol}} \approx 0.00208 \text{ mol} \)

### Step 3: Determine Stoichiometry Requirement
According to the balanced equation, for every 3 moles of NaHCO₃, 1 mole of C₆H₈O₇ is needed:

- Required moles of NaHCO₃ for 0.00208 mol C₆H₈O₇:
\[ 3 \times 0.00208 \text{ mol} = 0.00624 \text{ mol NaHCO}_3 \]

### Step 4: Limiting Reactant Determination:
- Available moles of NaHCO₃ = 0.00595 mol
- Required moles of NaHCO₃ = 0.00624 mol

NaHCO₃ is not sufficient to react with the available C₆H₈O₇ in the sample because the amount we have (0.00595 moles) is less than required (0.00624 moles).

### Conclusion:
**Sodium bicarbonate (NaHCO₃) is the limiting reactant.** The reaction will stop when NaHCO₃ runs out, even though there will be some unused citric acid remaining.