What Is the Percent Yield of SO2 from CS2 in This Reaction?

[Koskesh]

Homework Statement

What is the % yield of 0.112g S02 obtained from the combination of 0.0781g of CS2 according to the reaction CS2 + 3O2 --> CO2 + 2SO2

Homework Equations

I believe it's Actual yield / Theoretical yield x 100% ?

The Attempt at a Solution

Not sure would 0.112g go under the AY? I don't now even know where to start.

 

[Bohrok]

Looks like you know what you're doing. The equation's right and the actual yield is .112 g SO₂. How do you find the theoretical yield?

 

[Blueshift5]

Hello,

To calculate the percent yield, you will need to first calculate the theoretical yield of SO2 using stoichiometry. This can be done by setting up a mole ratio between CS2 and SO2. In this case, the mole ratio is 1:2, meaning for every 1 mole of CS2, 2 moles of SO2 are produced.

First, convert the given mass of CS2 (0.0781g) to moles by dividing by the molar mass of CS2 (76.14 g/mol). This gives you 0.001026 moles of CS2.

Next, use the mole ratio to determine the moles of SO2 produced. Since the ratio is 1:2, this means that for 0.001026 moles of CS2, 0.002052 moles of SO2 are produced.

Now, convert the moles of SO2 to grams by multiplying by the molar mass of SO2 (64.06 g/mol). This gives you a theoretical yield of 0.1314g of SO2.

To calculate the percent yield, divide the actual yield (0.112g) by the theoretical yield (0.1314g) and multiply by 100%. This gives you a percent yield of approximately 85.3%.

I hope this helps! Let me know if you have any further questions.