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)
GIVEN:
- 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
MY ATTEMPT:
(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.
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.
Ah, I see. Thank you!
I think you didn't use the M of the H2SO4.
Your calculations are almost correct, but there is a small error in your conversion factor. Let's go through the correct calculation step by step:
1. Convert the volume of H2SO4 to liters:
Given: 25.0 mL = 0.025 L
2. Use the molarity of H2SO4 to calculate the number of moles:
Given: 0.360 M H2SO4
moles of H2SO4 = 0.025 L H2SO4 * 0.360 mol/L H2SO4 = 0.009 mol H2SO4
3. Use stoichiometry to calculate the moles of NaHCO3 needed:
From the balanced equation, we see that 1 mole of H2SO4 reacts with 2 moles of NaHCO3.
moles of NaHCO3 = 0.009 mol H2SO4 * (2 mol NaHCO3 / 1 mol H2SO4) = 0.018 mol NaHCO3
4. Calculate the mass of NaHCO3 needed:
Given: Molar mass NaHCO3 = 84.01 g/mol
mass of NaHCO3 = 0.018 mol NaHCO3 * 84.01 g/mol = 1.512 g NaHCO3
So, the correct answer is that approximately 1.512 grams of NaHCO3 will be needed to neutralize the spilled H2SO4.