Question

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

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 $O_2$.

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8g}{32g/mole}=0.25moles$

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 $\frac{0.1}{2}=0.05$ 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 $CO$

$\text{ Mass of }CO=\text{ Moles of }CO\times \text{ Molar mass of }CO$

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

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