Consider the following balanced chemical equation: 
H₂O+ 2 MnO₄^– + 3 SO₃^2-  → 2 MnO₂ + 3 SO₄^2-+ 2 OH^–
How many grams of MnO₂ (MW:86.94 g/mol) will be created when 25.0 mL of 0.120 M MnO₄^– (MW:118.90 g/mol) reacts with 32.0 mL of 0.140 M SO₃^2- (MW:80.07 g/mol).  

• A. 0.073 g
• B. 0.089 g
• C. 0.19 g
• D. 0.26 g
• E. 0.67 g

Answer: (D)

Verified step by step guidance

1. Step 1: Begin by identifying the limiting reactant. Calculate the moles of MnO4– and SO32– using their respective concentrations and volumes. Use the formula: \( \text{moles} = \text{concentration} \times \text{volume} \).
2. Step 2: For MnO4–, use the concentration (0.120 M) and volume (25.0 mL). Convert the volume from mL to L by dividing by 1000. Calculate the moles of MnO4–.
3. Step 3: For SO32–, use the concentration (0.140 M) and volume (32.0 mL). Convert the volume from mL to L by dividing by 1000. Calculate the moles of SO32–.
4. Step 4: Use the stoichiometry of the balanced equation to determine which reactant is the limiting reactant. The balanced equation shows that 2 moles of MnO4– react with 3 moles of SO32–. Compare the mole ratio of the reactants to find the limiting reactant.
5. Step 5: Once the limiting reactant is identified, use its moles to calculate the moles of MnO2 produced using the stoichiometry of the balanced equation. Then, convert moles of MnO2 to grams using its molar mass (86.94 g/mol) with the formula: \( \text{mass} = \text{moles} \times \text{molar mass} \).