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Paul [167]
3 years ago
9

A sulfuric acid solution containing 571.3 g of h2so4 per liter of aqueous solution has a density of 1.329 g/cm3. Part a calculat

e the mass percentage of h2so4 in this solution. Express your answer using four significant figures. G
Chemistry
1 answer:
loris [4]3 years ago
3 0

Mass percentage of a solution is the amount of solute present in 100 g of the solution.

Given data:

Mass of solute H2SO4 = 571.3 g

Volume of the solution = 1 lit = 1000 ml

Density of solution = 1.329 g/cm3 = 1.329 g/ml

Calculations:

Mass of the given volume of solution = 1.329 g * 1000 ml/1 ml = 1329 g

Therefore we have:

571.3 g of H2SO4 in 1329 g of the solution

Hence, the amount of H2SO4 in 100 g of solution= 571.3 *100/1329 = 42.987

Mass percentage of H2SO4 (%w/w) is 42.99 %

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Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make ___ mol ammonium sulfate
Bingel [31]

Answer:

1) Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make 0.157 mol ammonium sulfate when you neutralize 11.00 g ammonium hydroxide.

2) 2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O.

Explanation:

  • Firstly, we should balance the equation of heptane combustion.
  • We can balance the equation by applying the conservation of mass to the equation.
  • The balanced equation is: <em>2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O.</em>
  • This means that every 2.0 moles of ammonium hydroxide (NH₄OH) will produce 1.0 mole of ammonium sulfate (NH₄)₂SO₄ when it is neutralized by sulfuric acid.
  • We need to calculate the no. of moles in 11.0 g of ammonium hydroxide that is neutralized using the relation: <em>n = mass/molar mass. </em>

n of 11.0 g of ammonium hydroxide (NH₄OH) = mass/molar mass = (11.0 g)/(35.04 g/mol) = 0.314 mol.

<u><em>Using cross multiplication: </em></u>

2.0 moles of NH₄OH make → 1.0 mole of (NH₄)₂SO₄.

0.314 mol of NH₄OH make → ??? moles of (NH₄)₂SO₄.

∴ The no. of moles of (NH₄)₂SO₄ that will be made from neutralizing (11.0 g) of NH₄OH = (0.314 mol)(1.0 mol)/(2.0 mol) = 0.157 mol.

<em>∴ Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make </em><em>0.157</em><em> mol ammonium sulfate when you neutralize 11.00 g ammonium hydroxide.</em>

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