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UNO [17]
3 years ago
11

Calculate the ratio of moles of H2O to moles of anhydrous CuSO4 in CuSO4.

Chemistry
1 answer:
rjkz [21]3 years ago
7 0

Answer:

CuSO₄.H₂O = 1

CuSO₄.3H₂O = 3

CuSO₄.5H₂O = 5

CuSO₄7.H₂O = 7

CuSO₄.9H₂O = 9

Explanation:

Some salts as CuSO₄ are presented in the hydratated form to give some stability in their uses.

Ratio of moles represents moles of H₂O / moles of CuSO₄.

In CuSO₄.H₂O you have 1 mole of water per mole of CuSO₄, Ratio is 1/1 = 1.

For CuSO₄.3H₂O are 3 moles of water per mole of CuSO₄. Ratio is 3/1 = 3

For CuSO₄.5H₂O are 5 moles of water per mole of CuSO₄. Ratio is 5/1 = 5

For CuSO₄.7H₂O are 7 moles of water per mole of CuSO₄. Ratio is 7/1 = 7

For CuSO₄.9H₂O are 9 moles of water per mole of CuSO₄. Ratio is 9/1 = 9

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Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of
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Answer:

  • <u>79%</u>

Explanation:

<u>1) Balanced chemical equation:</u>

  • 2S + 3O₂ → 2SO₃

<u>2) Mole ratio:</u>

  • 2 mol S : 3 mol O₂ : 2 mol SO₃

<u>3) Limiting reactant:</u>

  • Number of moles of O₂

        n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂

  • Number of moles of S:

         n = 7.0 g / 32.065 g/mol = 0.2183 mol S

  • Ratios:

        Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859

        Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5

Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.

<u>4) Calcuate theoretical yield (using the limiting reactant):</u>

  • 0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃

  • x = 0.1875 × 2 / 3 mol SO₃ =  0.125 mol SO₃

<u>5) Yield in grams:</u>

  • mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol =  10.0 g

<u>6) </u><em><u>Percent yield:</u></em>

  • Percent yield, % = (actual yield / theoretical yield) × 100
  • % = (7.9 g / 10.0 g) × 100 = 79%
6 0
3 years ago
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