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. Determine the moles of MnO4– using its concentration and volume. Use the formula: moles = concentration (M) × volume (L). Convert 25.0 mL to liters by dividing by 1000.
2. Determine the moles of SO32- using its concentration and volume. Use the formula: moles = concentration (M) × volume (L). Convert 32.0 mL to liters by dividing by 1000.
3. 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 required moles of each reactant to fully react with the other and determine which is the limiting reactant.
4. Use the stoichiometry of the balanced equation to find the moles of MnO2 produced. According to the equation, 2 moles of MnO4– produce 2 moles of MnO2. Use the moles of the limiting reactant to calculate the moles of MnO2.
5. Convert the moles of MnO2 to grams using its molar mass (86.94 g/mol). Use the formula: mass = moles × molar mass.