Question

Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 1.74 g of butane is mixed with 11. g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Answer : The mass of
`CO_2` produced will be, 5.3 grams.

Explanation : Given,

Mass of
`C_4H_(10)` = 1.74 g

Mass of
`O_2` = 11 g

Molar mass of
`C_4H_(10)` = 58 g/mole

Molar mass of
`O_2` = 32 g/mole

Molar mass of
`CO_2` = 44 g/mole

First we have to calculate the moles of
`C_4H_(10)` and
`O_2`.

`\text{Moles of }C_4H_(10)=\frac{\text{Mass of }C_4H_(10)}{\text{Molar mass of }C_4H_(10)}=(1.74g)/(58g/mole)=0.03moles`

`\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=(11g)/(32g/mole)=0.34moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`2C_4H_(10)+13O_2\rightarrow 8CO_2+10H_2O`

From the balanced reaction we conclude that

As, 2 moles of
`C_4H_(10)` react with 13 mole of
`O_2`

So, 0.03 moles of
`C_4H_(10)` react with
`(13)/(2)* 0.03=0.195` 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_4H_(10)` is a limiting reagent and it limits the formation of product.

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

As, 2 moles of
`C_4H_(10)` react to give 8 moles of
`CO_2`

So, 0.03 moles of
`C_4H_(10)` react to give
`(8)/(2)* 0.03=0.12` 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=(0.12mole)* (44g/mole)=5.3g`

Therefore, the mass of
`CO_2` produced will be, 5.3 grams.