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. First, determine the moles of MnO4– and SO32– using the molarity and volume of each solution. Use the formula: moles = molarity × volume (in liters).
2. Convert the volume from milliliters to liters for both MnO4– and SO32–. For MnO4–, 25.0 mL is 0.025 L, and for SO32–, 32.0 mL is 0.032 L.
3. Calculate the moles of MnO4–: moles = 0.120 M × 0.025 L. Similarly, calculate the moles of SO32–: moles = 0.140 M × 0.032 L.
4. Identify the limiting reactant by comparing the mole ratio from the balanced equation. The balanced equation shows that 2 moles of MnO4– react with 3 moles of SO32–. Calculate the ratio of moles available to determine the limiting reactant.
5. Using the moles of the limiting reactant, calculate the moles of MnO2 produced using the stoichiometry from the balanced equation. Then, convert moles of MnO2 to grams using its molar mass (86.94 g/mol) with the formula: grams = moles × molar mass.