If the typical percent yield of SO2 from the chemical reaction shown below is 75%, how much CS2 should be used in a reaction to ensure that a minimum of 15.0 g SO2 is produced?

11.875g of CS2

Explanation:

First, we try and get the reaction being talked about as the question doesn’t seem complete. From the question, we can see that that CS2 is used in the production of SO2. So let’s find this reaction!

Yes, the probable reaction is below;

CS2 + CO2 → 3C + 2 SO2

Carbon sulphide is reacting with carbon iv oxide to yield elemental solid carbon and sulphur iv oxide.

Let’s proceed to answer the question now!

Since 15.0g of SO2 is to be produced, with a percentage yield of 75%, the theoretical yield of SO2 will be;

15 * 100/75 = 20g

We can convert this value to moles by using the molar mass of SO2. The Molar mass of SO2 is 32 + 2(16) = 64g/mol

The number of moles of SO2 is thus 20/64 = 0.3125moles

We now proceed to get the number of moles of CS2.

From the reaction,

1 mole CS2 yielded 2 moles SO2

x mole CS2 will yield 0.3125 moles SO2

This x = 0.3125/2 = 0.15625 mole CS2

We then use the molar mass of CS2 to get the actual amount in grammes

The molar mass of CS2 is 12 + 32(2) = 12 + 64 = 76g/mol

The mass is thus 76 * 0.15625 = 11.875g

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