What Is The Theoretical Yield Of So3 Produced By The Quantities Described In Part A

What is the theoretical yield of SO3 produced by the quantities described in Part 1?

1.614x10^23 O2 molecules / 6.022x10^23 molecules/mole = 0.268 moles O2

5.72 g S / 32.065 g/mole = 0.162 mole S

2S + 3O2 --> 2SO3
0.162 mole S requires 0.243 mole O2 (ratio S : O2 = 2 : 3)
You have an excess of O2; S is limiting

0.162 mole S yields 0.162 mole SO3
0.162 mole x 80.06 g/mole = 12.97 g SO3
ANSWER: 13.0 g SO3

What is the theoretical yield of produced by the quantities described in Part A? ?

Moles O2 = 8.39 x 10^22 / 6.02 x 10^23 = 0.139
Moles S = 2.98 g / 32.066 g/mol = 0.0929

2 : 3 = 0.0929 : x
x = 0.139 moles O2 needed
no reactant is in excess

we wuold get 0.0929 moles of SO3
mass = 0.0929 mol x 80 g/mol =7.43 g

What is the theoretical yield of SO3 produced by the quantities described in Part A?

Contd from the previous que........
2S + 3O2 ---> 2SO3

Mw of SO3 = 32 + 3(16) = 32 + 48 = 80g
2moles = 2 x 80 = 160g

64g of S2 yields 160g of SO3
4.22g of S2 yields ?
? = [ 4.22 x 160 ] / 64
? = 675.2 / 64 = 10.55g

Theoretical yield of SO3 = 10.55g

What is the theoretical yield of SO3 produced by the quantities described in Part A?

First write the general reaction out:

S + O2 -> SO3
Now balance both sides:
2S + 3O2 -> 2SO3
Great, now using the 5.51 grams of Sulfur, we need to figure out how much SO3 this will produce.

First, Sulfur has an atomic mass of ~32.065 g/mol. Now lets solve for how many mols of S we have.
5.51 g * 1 mol / 32.065 g = 0.1718 mols of sulfur.
Finally, we figure out how many moles of SO3 can be produced by 0.1718 moles of sulfur. To do this, just multiply and divide by the stoichiometric coefficients between the two.
0.1718 mols of sulfur * 2 mol SO3 / 2 mol S = 0.1718 moles of SO3

There it is, 0.1718 moles of SO3 is the theoretical yield. Hope this helps!

What is the theoretical yield of SO3 produced by quantities described in part a?

2 S + 3 O2 → 2 SO3

(8.48 g S) / (32.0655 g S/mol) x (2 mol SO3 / 2 mol S) x (80.0638 g SO3/mol) =
21.2 g SO3

Stoichiometry : Elemental S reacts with O2 to form SO3 according to the reaction?

Elemental S reacts with O2 to form SO3 according to the reaction
2S+3O2→2SO3

Part A
How many O2 molecules are needed to react with 7.49g of S?
Express your answer numerically in units of molecules.

Part B
What is the theoretical yield of SO3 produced by the quantities described in Part A?
Express your answer numerically in grams.

Elemental S reacts with O2 to form SO3 according to the reaction 2S + 3O2 = 2SO3?

Erm, I believe the equation is incorrect. If sulphur reacted with oxygen, sulphur dioxide would be produced and when further reacted with oxygen, produces sulphur trioxide. The overall equation is actually SO2+S+2O2=2SO3.

A- With that in mind, we need to know the number of moles present in 2.38 g of S. Now, the Relative Atomic Mass(RAM) of S is 32. Thus, the number of moles of S present would be 2.38/32= 0.074375. According to the equation, the number of moles of O2 is twice that of S, thus the number of moles of O2= 0.14875. 1 mole of O2 has 6.02x10^23 molecules, thus 0.14875 moles of O2 have 8.95475x10^22 molecules.

B- The number of moles of SO3 yielded is equal to that of S required. The Relative Molecular Mass(RMM) of SO3 is 80. Therefore, the theoretical yield of SO3 would be 80x0.074375= 5.95g.

Hope this helps!!

Elemental S reacts with O2 to form SO3 according to the reaction 2S+3O2→2SO3?

number of moles of S = mass/molar mass = 4,61/32 = 0,144 moles
3/2x as much moles of oxygen react with 0,144 moles of sulphur = 3/2 x 0,144 = 0,216 moles of oxygen
number of oxygen atoms = number of moles x Avogadro constant = 0,216 x (6,023 x 10 to the power of 23) = (1,3 x 10 to the power of 23) molecules of oxygen

Theoretical yield:
number of moles of SO3 formed = number of moles of S = 0,144
mass of SO3 formed = number of moles of SO3 x molar mass = 0,144 x 80 = 11,52 grams