What is the theoretical yield of so3 produced by the quantities described in part a?\?
1 answer:
The problem here is incomplete. Luckily, I found a similar problem from another website which is shown in the attached picture. So, the useful information here are the stoichiometric coefficients of the reaction and the initial amount of 7.49 g of S. The solution is as follows:
Molar mass of S = 32.06 g/mol
Molar mass of SO₃ = 80.06 g/mol
Theo yield of SO₃ = (7.49 g S)(1 mol S/32.06 g)(2 mol SO₃/2 mol S)(80.06 g SO₃/mol) = <em>18.7 g SO₃</em>
Molar mass of S = 32.06 g/mol
Molar mass of SO₃ = 80.06 g/mol
Theo yield of SO₃ = (7.49 g S)(1 mol S/32.06 g)(2 mol SO₃/2 mol S)(80.06 g SO₃/mol) = <em>18.7 g SO₃</em>
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Correct answer: Option D, <span>K = 5.04 × 10^52
</span>
Reason:
We know that,
Ecell =
,
where n = number of electrons = 2 (in present case)
K = equilibrium constant.
Also, Ecell = <span>+1.56 v
Therefore, 1.56 = </span>
Therefore, log (K) = 52.703
Therefore, K = 5.04 X 10^52
</span>
Reason:
We know that,
Ecell =
where n = number of electrons = 2 (in present case)
K = equilibrium constant.
Also, Ecell = <span>+1.56 v
Therefore, 1.56 = </span>
Therefore, log (K) = 52.703
Therefore, K = 5.04 X 10^52