Answer to Question #253932 in General Chemistry for jacob

Question

If 32 mL

mL of 5.4 M

M H

2

SO

4

H2SO4 was spilled, what is the minimum mass of NaHCO

3

NaHCO3 that must be added to the spill to neutralize the acid?

Answer 1

Given Molarity of H₂SO₄ = 5.4 M

Volume of H₂SO₄ = 32 mL

the balanced chemical equation of the reaction will be

H₂SO₄ + 2NaHCO₃ --> Na₂SO₄ + 2CO₂ + 2H₂O

Molarity = (Moles * 1000) / Volume

5.4 = ((Moles of H₂SO₄ ) * 1000) / 32

Moles of H₂SO₄ = (5.4 * 32)/1000 = 0.1728 moles

Now from balanced chemical reaction, we can see that 1 mole of H₂SO₄ reacts with 2 moles of NaHCO₃

Hence, moles of NaHCO₃ = 2 *0.1728 = 0.3456 moles

Molecular mass of NaHCO₃ = 23 + 1 + 12 + (3*16) = 84 g/mole

So, mass of NaHCO₃ = Moles NaHCO₃ * Molecular mass

mass of NaHCO₃ = 84 * 0.3456 = 29g

So the minimum amount of NaHCO₃ that must be added to neutralize the acid is 29g.