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Alex_Xolod [135]
2 years ago
5

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
Chemistry
1 answer:
jek_recluse [69]2 years ago
8 0

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.

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First we have to calculate the charge.

Formula used : Q=I\times t

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Now put all the given values in this formula, we get

Q=10A\times 600s=6000C

Now we have to calculate the number of atoms deposited.

As, 1 atom require charge to deposited = 2\times (1.6\times 10^{-19})  

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n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

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