Given that the reaction between sodium bicarbonate (NaHCO_3) and sulfuric acid (H_2SO_4) is: 2 NaHCO_3 + H_2SO_4 → Na_2SO_4 + 2 CO_2 + 2 H_2O, how many moles of NaHCO_3 are required to completely neutralize 0.8 mL of H_2SO_4 (density = 1.84 g/mL, molar mass = 98.08 g/mol) at STP?

• A. 0.03 × 10^{-3} mol
• B. 0.03 mol
• C. 0.08 mol
• D. 0.03 mmol

Answer: (A)

Verified step by step guidance

1. Calculate the mass of H_2SO_4 using its volume and density with the formula: \(\text{mass} = \text{density} \times \text{volume}\). Here, volume should be in mL and density in g/mL to get mass in grams.
2. Convert the mass of H_2SO_4 to moles using its molar mass with the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\).
3. Use the balanced chemical equation to find the mole ratio between NaHCO_3 and H_2SO_4. According to the equation, 2 moles of NaHCO_3 react with 1 mole of H_2SO_4.
4. Calculate the moles of NaHCO_3 required by multiplying the moles of H_2SO_4 by the mole ratio (2:1).
5. Express the final answer in appropriate units (moles or millimoles) as required.