A beaker containing 25.0 mL of 0.360 M H2SO4 spills on the counter. How much baking soda, NaHCO3, will be needed to neutralize the acid?

H2SO4 (aq) + 2 NaHCO3 (s) -> Na2SO4 (aq) + 2 H2CO3 (l)

- I've converted 25.0 mL to 0.025 L H2SO4.
- 0.360 M H2SO4
- Molar mass H2SO4: 98.08 g/mol
- Molar mass NaHCO3: 84.01 g/mol

(0.025 L H2SO4 x 1 mol H2SO4 x 1 mol NahCO3 x 84.01 g NaHCO3) / (1 mol H2SO4 x 2 mol NaHCO3 x 84.01 NaHCO3 x 1 mole NahCO3)
= 0.0125 g NaHCO3

I'm not sure if I'm right.

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  1. mols H2SO4 = M x L =?
    mols NaHCO3 = 2x that
    grams NaHCO3 = mols NaHCO3 x molar mass NaHCO3.
    I think the answer is closer to 1.5 grams.

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  2. Ah, I see. Thank you!

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  3. I think you didn't use the M of the H2SO4.

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