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Part B

What is the percent yield of carbon disulfide if the reaction of 35.9 g of sulfur dioxide produces 12.7 g of carbon disulfide?

Express your answer with the appropriate units.

Answer :

Final answer:

The percent yield of carbon disulfide is 29.7%.


Explanation:

The percent yield can be calculated using the formula:

Percent yield = (Actual yield / Theoretical yield) x 100

To find the percent yield of carbon disulfide, we need to determine the theoretical yield. The balanced chemical equation for the reaction is:

2 SO2 + C → 2 CS2 + 2 O

From the equation, we can see that 2 moles of sulfur dioxide (SO2) react with 1 mole of carbon (C) to produce 2 moles of carbon disulfide (CS2). We can use the molar mass of sulfur dioxide (64.07 g/mol) to calculate the moles of sulfur dioxide:

35.9 g SO2 x (1 mol SO2 / 64.07 g SO2) = 0.5608 mol SO2

Using the stoichiometric ratio from the balanced equation, we can determine the moles of carbon disulfide:

0.5608 mol SO2 x (2 mol CS2 / 2 mol SO2) = 0.5608 mol CS2

Finally, we can convert the moles of carbon disulfide to grams:

0.5608 mol CS2 x (76.14 g CS2 / 1 mol CS2) = 42.74 g CS2

Now, we can calculate the percent yield:

Percent yield = (12.7 g / 42.74 g) x 100 = 29.7%


Learn more about Percent Yield here:

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