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A 5.37 g sample of household cleaner is dissolved in 25.0 ml of water to form NH₄OH. This solution required 37.3 cm3 of 0.360 M sulphuric acid for neutralization. Calculate the percentage mass of ammonia (NH₃) in the cleaner? 2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O

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To calculate the percentage mass of ammonia (NH₃) in the cleaner, we need to use the balanced chemical equation provided and the information given.

The balanced chemical equation is:

2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O

From the equation, we can see that the ratio between NH₄OH and H₂SO₄ is 2:1.

1. Calculate the number of moles of H₂SO₄ used:

- Volume of H₂SO₄ solution = 37.3 cm³ = 0.0373 L

- Molarity of H₂SO₄ solution = 0.360 M

- Moles of H₂SO₄ = Molarity x Volume = 0.360 mol/L x 0.0373 L = 0.0134 mol

2. Since the ratio between NH₄OH and H₂SO₄ is 2:1, the number of moles of NH₄OH used is also 0.0134 mol.

3. Calculate the molar mass of NH₄OH:

- NH₄OH = 1(N) + 4(H) + 1(O) = 14.01 g/mol + 4.03 g/mol + 16.00 g/mol = 34.04 g/mol

4. Calculate the mass of NH₄OH used:

- Mass = Moles x Molar mass = 0.0134 mol x 34.04 g/mol = 0.454 g

5. Calculate the percentage mass of NH₃ in the cleaner:

- Percentage mass = (Mass of NH₃ / Mass of cleaner) x 100%

- The mass of NH₃ is twice the mass of NH₄OH used (from the balanced chemical equation).

- Mass of NH₃ = 2 x Mass of NH₄OH used = 2 x 0.454 g = 0.908 g

- Mass of cleaner = 5.37 g (given)

- Percentage mass of NH₃ = (0.908 g / 5.37 g) x 100% ≈ 16.9%

Therefore, the percentage mass of ammonia (NH₃) in the cleaner is approximately 16.9%.

User Ryan Kinal
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