Question

Consider the reaction of C3H8 with O2 to form CO2 and H2O. If 5.11 g O2 is reacted with excess C3H8 and 3.35 g of CO2 is ultimately isolated, what is the percent yield for the reaction? Percent yield = %

Answer 1

Answer : The percent yield of the reaction is, 79.8 %

Explanation : Given,

Mass of
`O_2` = 5.11 g

Molar mass of
`O_2` = 32 g/mole

Molar mass of
`CO_2` = 44 g/mole

First we have to calculate the moles of
`O_2`.

`\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=(5.11g)/(32g/mole)=0.159mole`

Now we have to calculate the moles of
`CO_2`.

The balanced chemical reaction will be,

`C_3H_8+5O_2\rightarrow 3CO_2+4H_2O`

From the balanced reaction, we conclude that

As, 5 moles of
`O_2` react to give 3 moles of
`CO_2`

So, 0.159 moles of
`O_2` react to give
`(3)/(5)* 0.159=0.0954` moles of
`CO_2`

Now we have to calculate the mass of
`CO_2`

`\text{Mass of }CO_2=\text{Moles of }CO_2* \text{Molar mass of }CO_2`

`\text{Mass of }CO_2=(0.0954mole)* (44g/mole)=4.1976g`

The theoretical yield of
`CO_2` = 4.1976 g

The actual yield of
`CO_2` = 3.35 g

Now we have to calculate the percent yield of
`CO_2`

`\%\text{ yield of }CO_2=\frac{\text{Actual yield of }CO_2}{\text{Theoretical yield of }CO_2}* 100=(3.35g)/(4.1976g)* 100=79.8\%`

Therefore, the percent yield of the reaction is, 79.8 %