Question

Give the theoretical yield, in grams, of CO2 from the reaction of 4.000 moles of C8H18 with 4.000 moles of 02 2 C8H18 25 02 16 CO2+ 18 H20

Answer 1

Answer : The theoretical yield of
`CO_2` is, 112.64 grams.

Explanation : Given,

Given moles of
`C_8H_(18)` = 4 moles

Given moles of
`O_2` = 4 moles

Molar mass of
`CO_2` = 44 g/mole

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`2C_8H_(18)+25O_2\rightarrow 16CO_2+18H_2O`

From the given balanced reaction, we conclude that

As, 25 moles of
`O_2` react with 2 moles of
`C_8H_(18)`

So, 4 moles of
`O_2` react with
`(2)/(25)* 4=0.32` moles of
`C_8H_(18)`

From this we conclude that,
`C_8H_(18)` is an excess reagent because the given moles are greater than the required moles and
`O_2` is a limiting reagent because it limits the formation of product.

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

As, 25 moles of
`O_2` react to give 16 moles of
`CO_2`

So, 4 moles of
`O_2` react to give
`(16)/(25)* 4=2.56` 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=(2.56mole)* (44g/mole)=112.64g`

Therefore, the theoretical yield of
`CO_2` is, 112.64 grams.