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:
Part 1: Neutralizing the Spill with Sodium Bicarbonate
The balanced chemical equation for the reaction between sulfuric acid (H₂SO₄) and sodium bicarbonate (NaHCO₃) is: H2SO4(aq)+2NaHCO3(s)→Na2SO4(aq)+2H2O(l)+2CO2(g)\text{H}_2\text{SO}_4(aq) + 2\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2\text{H}_2\text{O}(l) + 2\text{CO}_2(g)H2SO4(aq)+2NaHCO3(s)→Na2SO4(aq)+2H2O(l)+2CO2(g)
Step 1: Determine Moles of Sulfuric Acid
We are given 2.48 L of 3.00 M sulfuric acid. Moles of H2SO4=2.48 L×3.00 mol/L=7.44 mol of H2SO4\text{Moles of H}_2\text{SO}_4 = 2.48 \, \text{L} \times 3.00 \, \text{mol/L} = 7.44 \, \text{mol of H}_2\text{SO}_4Moles of H2SO4=2.48L×3.00mol/L=7.44mol of H2SO4
Step 2: Determine Moles of Sodium Bicarbonate Available
The molar mass of NaHCO₃ is 84.01 g/mol. We are given 1.25 kg of NaHCO₃, or 1250 g. Moles of NaHCO3=1250 g84.01 g/mol≈14.87 mol of NaHCO3\text{Moles of NaHCO}_3 = \frac{1250 \, \text{g}}{84.01 \, \text{g/mol}} \approx 14.87 \, \text{mol of NaHCO}_3Moles of NaHCO3=84.01g/mol1250g≈14.87mol of NaHCO3
Step 3: Calculate the Required Moles of NaHCO₃
From the balanced equation, 1 mole of H₂SO₄ reacts with 2 moles of NaHCO₃. For 7.44 moles of H₂SO₄, the required moles of NaHCO₃ are: 7.44 mol H2SO4×2 mol NaHCO3/mol H2SO4=14.88 mol NaHCO37.44 \, \text{mol H}_2\text{SO}_4 \times 2 \, \text{mol NaHCO}_3/\text{mol H}_2\text{SO}_4 = 14.88 \, \text{mol NaHCO}_37.44mol H2SO4×2mol NaHCO3/mol H2SO4=14.88mol NaHCO3
Step 4: Compare the Available NaHCO₃ with the Required Amount
We have 14.87 mol of NaHCO₃ available, but we need 14.88 mol to completely neutralize the sulfuric acid.
The available NaHCO₃ is slightly less than needed, but the difference is very small. We can estimate the percent difference as: 14.88−14.8714.88×100≈0.07%\frac{14.88 – 14.87}{14.88} \times 100 \approx 0.07\%14.8814.88−14.87×100≈0.07%
This difference is insignificant, so there is effectively enough NaHCO₃ to neutralize the sulfuric acid.
Answer:
The correct answer is c. Yes, and there is about 10% more NaHCO₃ than needed.
Part 2: 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)
Step 1: Identify Oxidation States
- In H₃AsO₃ (arsenous acid), arsenic (As) has an oxidation state of +3.
- In H₃AsO₄ (arsenic acid), arsenic (As) has an oxidation state of +5.
- In KBrO₃, bromine (Br) has an oxidation state of +5.
- In KBr, bromine (Br) has an oxidation state of -1.
Step 2: Determine Oxidation and Reduction
- Oxidation: The element that increases in oxidation state is being oxidized. In this case, arsenic goes from +3 to +5, so arsenic (As) is oxidized.
- Reduction: The element that decreases in oxidation state is being reduced. In this case, bromine goes from +5 to -1, so bromine (Br) is reduced.
Answer:
- The element that is oxidized is As (arsenic).
- The element that is reduced is Br (bromine).
