Question

When 0.100 mol of carbon is burned in a closed vessel with8.00

g of oxygen, how many grams of carbon dioxide can form?

Answer 1

Answer : The mass of carbon monoxide form can be 2.8 grams.

Solution : Given,

Moles of C = 0.100 mole

Mass of
`O_2` = 8.00 g

Molar mass of
`O_2` = 32 g/mole

Molar mass of CO = 28 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}=(8g)/(32g/mole)=0.25moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`2C+O_2\rightarrow 2CO`

From the balanced reaction we conclude that

As, 2 mole of
`C` react with 1 mole of
`O_2`

So, 0.1 moles of
`C` react with
`(0.1)/(2)=0.05` 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` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`CO`

From the reaction, we conclude that

As, 2 mole of
`C` react to give 2 mole of
`CO`

So, 0.1 moles of
`C` react to give 0.1 moles of
`CO`

Now we have to calculate the mass of
`CO`

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

`\text{ Mass of }CO=(0.1moles)* (28g/mole)=2.8g`

Therefore, the mass of carbon monoxide form can be 2.8 grams.