Question

Sulfur hexafluoride is produced by reacting elemental sulfur with fluorine gas. S8(s) + 24 F2(g) LaTeX: \longrightarrow⟶ 8 SF6(g) What is the percent yield if 125 moles SF6 is isolated from the reaction of 1000 moles F2 and 40 moles sulfur?

Answer 1

Answer:

39.1%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

S8(s) + 24F2(g) —> 8SF6(g)

Next, we shall determine the limiting reactant. This is illustrated below:

From the balanced equation above,

1 mole of S8 reacted with 24 moles of F2.

Therefore, 40 moles of S8 will react with = 40 x 24 = 960 moles of F2.

Now we can see that it will take a lesser amount of F2 than what was given to react completely with 40 moles of S8. Therefore, S8 is the limiting reactant and F2 is the excess reactant.

Next, we shall determine the theoretical yield of SF6.

Here we shall be using the limiting reactant because all of it is used up in the reaction and it will also produce the maximum yield of SF6.

The limiting reactant is S8 and the theoretical yield of SF6 is obtained as follow:

From the balanced equation above,

1 mole of S8 produced 8 moles of SF6.

Therefore, 40 moles of S8 will produce = 40 x 8 = 320 moles of SF6.

Therefore, the theoretical yield of SF6 is 320 moles

Finally, the percentage yield of SF6 can be calculated as follow:

Actual yield of SF6 = 125 moles

Theoretical yield of SF6 = 320

Percentage yield of SF6 =..?

Percentage yield = Actual yield /Theoretical yield x100

Percentage yield = 125/320 x 100

Percentage yield = 39.1%

Therefore, the percentage yield of SF6 is 39.1%

Answer 2

Answer:

39%

Explanation:

We must first determine the limiting reactant. The limiting reactant will give the least number of moles of product.

For S8

From the reaction equation;

1 mole of S8 yields 8 moles of SF6

40 moles of sulphur yields 40 ×8/1 = 320 moles of SF6

For F2

24 moles of F2 yields 8 moles of SF6

1000 moles of F2 yields 1000 × 8/24 = 333.3 moles of SF6

Hence S8 is the limiting reactant

Using the theoretical yield of the limiting reactant;

Thus;

% yield= actual yield / theoretical yield × 100/1

% yield = 125/320 × 100/1 = 39%