Question

This requires you to use all the things we have learned in this chapter. 1.55 grams of Carbon disulfide

is added to 2.83 grams of Oxygen to form Carbon dioxide and Sulfur dioxide. If the reaction yields
1.1 grams of the Sulfur dioxide what was the percent yield?
27%
14%
Ο Ο Ο Ο Ο
74%
42%
O 63%

Answer 1

Answer: The percent yield is, 42%

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of carbon disulphide}=(1.55g)/(76g/mol)=0.020moles`

`\text{Moles of oxygen}=(2.83g)/(32g/mol)=0.088moles`

The balanced chemical reaction is:

`CS_2+3O_2(g)\rightarrow CO_2+2SO_2`

According to stoichiometry :

1 moles of
`CS_2` require = 3 moles of
`O_2`

Thus 0.020 moles of
`CS_2` will require=
`(3)/(1)* 0.020=0.060moles` of
`O_2`

Thus
`CS_2` is the limiting reagent as it limits the formation of product and
`O_2` is the excess reagent.

As 1 mole of
`CS_2` give = 2 moles of
`SO_2`

Thus 0.020 moles of
`CS_2` give =
`(2)/(1)* 0.020=0.040moles` of
`SO_2`

Theoretical mass of
`SO_2=moles* {\text {Molar mass}}=0.040moles* 64g/mol=2.56g`

Actual mass of
`SO_2` = 1.1 g

Now we have to calculate the percent yield

`\%\text{ yield}=\frac{\text{Actual yield }}{\text{Theoretical yield}}* 100=(1.1g)/(2.56g)* 100=42\%`

Therefore, the percent yield is, 42%