Question

Problem Page Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 28. g of ethane is mixed with 190. g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits. Clears your work. Undoes your last action. Provides information about entering answers. g

Answer 1

Answer : The maximum mass of carbon dioxide produced by the chemical reaction can be, 82.3 grams.

Solution : Given,

Mass of
`C_2H_6` = 28 g

Mass of
`O_2` = 190 g

Molar mass of
`O_2` = 32 g/mole

Molar mass of
`C_2H_6` = 30 g/mole

Molar mass of
`CO_2` = 44 g/mole

First we have to calculate the moles of
`C_2H_6` and
`O_2`.

`\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=(28g)/(30g/mole)=0.933moles`

`\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=(190g)/(32g/mole)=5.94moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O`

From the balanced reaction we conclude that

As, 2 mole of
`C_2H_6` react with 7 mole of
`O_2`

So, 0.933 moles of
`C_2H_6` react with
`(7)/(2)* 0.933=3.26` moles of
`O_2`

From this we conclude that,
`O_2` is an excess reagent because the given moles are greater than the required moles and
`C_2H_6` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`CO_2`

From the reaction, we conclude that

As, 2 mole of
`C_2H_6` react to give 4 mole of
`CO_2`

So, 0.933 moles of
`C_2H_6` react to give
`(4)/(2)* 0.933=1.87` 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=(1.87moles)* (44g/mole)=82.3g`

Therefore, the maximum mass of carbon dioxide produced by the chemical reaction can be, 82.3 grams.