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2 years ago

If it takes 38.70cm of 1.90M NaOH to neutralize 10.30cm of H2SO4 in a battery, what is the molarity of H2SO4?

Chemistry

1 answer:

slamgirl [31]2 years ago

8 0

Answer:

The molarity of the acid, H₂SO₄ is 3.57 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ (nₐ) = 1

Mole ratio of the base, NaOH (n₆) = 2

Finally, we shall determine the molarity of the acid, H₂SO₄. This can be obtained as follow:

Volume of base, NaOH (V₆) = 38.70 cm³

Molarity of base, NaOH (M₆) = 1.90M

Volume of acid, H₂SO₄ (Vₐ) = 10.30 cm³

Molarity of acid, H₂SO₄ (Mₐ) =?

MₐVₐ / M₆V₆ = nₐ/n₆

Mₐ × 10.3 / 1.9 × 38.70 = 1/2

Mₐ × 10.3 / 73.53 = 1/2

Cross multiply

Mₐ × 10.3 × 2 = 73.53 × 1

Mₐ × 20.6 = 73.53

Divide both side by 20.6

Mₐ = 73.53 / 20.6

Mₐ = 3.57 M

Thus, the molarity of the acid, H₂SO₄ is 3.57 M

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$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

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Putting values in above equation, we get:

$\text{Moles of atmosphere}=\frac{5.00\times 10^{21}g}{28.96g/mol}=1.73\times 10^{20}mol$

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Moles of carbon dioxide in air = $\frac{0.0415}{100}\times 1.73\times 10^{20}=7.18\times 10^{16}mol$

Hence, the moles of oxygen and carbon dioxide in air is $3.63\times 10^{19}mol$ and $7.18\times 10^{16}mol$ respectively

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