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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The empirical formula is obtained by calculating the mole ratios of the atoms in the elements.
The number of moles =mass/ R.A.M
For hydrogen, no. of moles=8/1=8
For oxygen, no. Of moles=64/16=4
The tabular solution is attached.
The number of moles =mass/ R.A.M
For hydrogen, no. of moles=8/1=8
For oxygen, no. Of moles=64/16=4
The tabular solution is attached.