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) 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:

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

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.