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Licemer1 [7]
2 years ago
15

A 25.0 mL sample of sulfuric acid is completely neutralized by adding 32.8 mL of 0.116 mol/L ammonia solution. Ammonium sulfate

is formed. What is the concentration of the sulfuric acid?
Chemistry
1 answer:
Paul [167]2 years ago
7 0

Answer:

0.08 mol L-1

Explanation:

Sulfuric acid Formula: H2SO4

Ammonia Formula: NH3

Ammonium sulfate Formula: (NH₄)₂SO₄

H2SO4 + 2NH3 = 2NH4+ + SO4 2-

H2SO4 + 2NH3 = (NH₄)₂SO₄

H2SO4 = (1/2)x (32.8 x 10^-3 L x 0.116 mol L-1)/25 x 10^-3 L

= 0.08 mol L-1

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Rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

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Rate of disappearance of reactants = rate of appearance of products

                     ⇒ -\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}  -----------------------------(1)

    Given that the rate of disappearance of oxygen = -\frac{d[O_{2} ]}{dt} = 3.64 x 10⁻³ M/s

             So the rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = ?

from equation (1) we can write

                                   \frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]

                                ⇒ \frac{d[SO_{3}] }{dt} = 2 x 3.64 x 10⁻³ M/s

                                ⇒ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

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3 years ago
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