Question

Students working in lab accidentally spilled 17 l of 3.0 m h2so4 solution. they find a large container of acid neutralizer that contains baking soda, nahco3. how many grams of baking soda will be needed to neutralize the sulfuric acid spill? do not include a unit in your answer or it will be counted wrong. use correct significant figures

Answer 1

Answer is: 8568.71 of baking soda.

Balanced chemical reaction: H₂SO₄ + 2NaHCO₃ → Na₂SO₄ + 2CO₂ + 2H₂O.

V(H₂SO₄) = 17 L; volume of the sulfuric acid.

c(H₂SO₄) = 3.0 M, molarity of sulfuric acid.

n(H₂SO₄) = V(H₂SO₄) · c(H₂SO₄).

n(H₂SO₄) = 17 L · 3 mol/L.

n(H₂SO₄) = 51 mol; amount of sulfuric acid.

From balanced chemical reaction: n(H₂SO₄) : n(NaHCO₃) = 1 :2.

n(NaHCO₃) = 2 · 51 mol.

n(NaHCO₃) = 102 mol, amount of baking soda.

m(NaHCO₃) = n(NaHCO₃) · M(NaHCO₃).

m(NaHCO₃) = 102 mol · 84.007 g/mol.

m(NaHCO₃) = 8568.714 g; mass of baking soda.

Answer 2

The mass of baking soda required is 4284 g.

The reaction equations given as;

H2SO4(aq) + NaHCO3(aq) ------>  H2O(l) + CO2(g) + NaHSO4(aq)

The number of moles of H2SO4 that was spilled is obtained from;

n = 17 L × 3M = 51 moles

We can see from the reaction equation that the reaction is a 1:1 reaction, this implies that 51 moles of NaHCO3.

Hence, mass of baking soda required = 51 moles ×84 g/mol = 4284 g

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