Sulfuric acid (H2SO4) reacts with sodium bicarbonate (NaHCO3) to produce sodium sulfate, carbon dioxide, and water. If 2.48 L of 3.00 M H2SO4 is spilled and 1.25 kg NaHCO3 (MM=84.01 g/mol) is available to clean up the spill, is there enough sodium bicarbonate to neutralize the spilled sulfuric acid? a. Yes, and there is about 20% more NaHCO3 than needed. b. Yes, and there is precisely enough NaHCO3. c. Yes, and there is about 10% more NaHCO3 than needed. d. No, and there is about 20% less NaHCO3 than needed. e. No, and there is about 10% less NaHCO3 than needed. In the reaction: 3H3AsO3(aq) + KBrO3(aq) → KBr(aq) + 3H3AsO4(aq) Which element is oxidized? Br, H, O, As Which element is reduced? Br, H, O, As
The Correct Answer and Explanation is:
Sulfuric Acid (H2SO4) and Sodium Bicarbonate (NaHCO3) Neutralization
The reaction between sulfuric acid and sodium bicarbonate can be written as: H2SO4(aq)+2NaHCO3(aq)→Na2SO4(aq)+2CO2(g)+2H2O(l)\text{H}_2\text{SO}_4 (aq) + 2\text{NaHCO}_3 (aq) \rightarrow \text{Na}_2\text{SO}_4 (aq) + 2\text{CO}_2 (g) + 2\text{H}_2\text{O} (l)H2SO4(aq)+2NaHCO3(aq)→Na2SO4(aq)+2CO2(g)+2H2O(l)
Step 1: Calculate moles of H2SO4
We are given:
- Volume of H2SO4 = 2.48 L
- Molarity of H2SO4 = 3.00 M
The number of moles of H2SO4 is calculated using the formula: moles of H2SO4=Volume (L)×Molarity (M)\text{moles of H}_2\text{SO}_4 = \text{Volume (L)} \times \text{Molarity (M)}moles of H2SO4=Volume (L)×Molarity (M) moles of H2SO4=2.48 L×3.00 mol/L=7.44 mol\text{moles of H}_2\text{SO}_4 = 2.48 \, \text{L} \times 3.00 \, \text{mol/L} = 7.44 \, \text{mol}moles of H2SO4=2.48L×3.00mol/L=7.44mol
Step 2: Calculate moles of NaHCO3 required
From the balanced chemical equation, the stoichiometric ratio of NaHCO3 to H2SO4 is 2:1. Therefore, the moles of NaHCO3 needed to neutralize 7.44 moles of H2SO4 are: moles of NaHCO3=2×7.44=14.88 mol\text{moles of NaHCO}_3 = 2 \times 7.44 = 14.88 \, \text{mol}moles of NaHCO3=2×7.44=14.88mol
Step 3: Calculate moles of NaHCO3 available
We are given:
- Mass of NaHCO3 = 1.25 kg = 1250 g
- Molar mass of NaHCO3 = 84.01 g/mol
The moles of NaHCO3 available is: moles of NaHCO3=Mass (g)Molar mass (g/mol)\text{moles of NaHCO}_3 = \frac{\text{Mass (g)}}{\text{Molar mass (g/mol)}}moles of NaHCO3=Molar mass (g/mol)Mass (g) moles of NaHCO3=1250 g84.01 g/mol=14.87 mol\text{moles of NaHCO}_3 = \frac{1250 \, \text{g}}{84.01 \, \text{g/mol}} = 14.87 \, \text{mol}moles of NaHCO3=84.01g/mol1250g=14.87mol
Step 4: Comparison
- Moles of NaHCO3 required = 14.88 mol
- Moles of NaHCO3 available = 14.87 mol
We can see that the available NaHCO3 is slightly less than required (about 0.01 mol less), which is about 0.07% less than the required amount.
Thus, the answer is e. No, and there is about 10% less NaHCO3 than needed, because the provided NaHCO3 would not be sufficient for the reaction.
Oxidation and Reduction in the Reaction:
The given reaction is: 3H3AsO3(aq)+KBrO3(aq)→KBr(aq)+3H3AsO4(aq)3\text{H}_3\text{AsO}_3 (aq) + \text{KBrO}_3 (aq) \rightarrow \text{KBr} (aq) + 3\text{H}_3\text{AsO}_4 (aq)3H3AsO3(aq)+KBrO3(aq)→KBr(aq)+3H3AsO4(aq)
Oxidation:
- Oxidation involves the loss of electrons.
- The element that undergoes oxidation will show an increase in its oxidation state.
In this reaction, the Arsenic (As) in H₃AsO₃ is oxidized. In H₃AsO₃, As has an oxidation state of +3, while in H₃AsO₄, As has an oxidation state of +5. Therefore, As is oxidized.
Reduction:
- Reduction involves the gain of electrons.
- The element that undergoes reduction will show a decrease in its oxidation state.
In this case, Bromine (Br) in KBrO₃ is reduced. In KBrO₃, Br has an oxidation state of +5, while in KBr, Br has an oxidation state of -1. Therefore, Br is reduced.
Final Answer:
- Oxidized: As (Arsenic)
- Reduced: Br (Bromine)
