Respuesta :
Answer : The mass of carbon monoxide form can be 2.8 grams.
Solution : Given,
Moles of C = 0.100 mole
Mass of [tex]O_2[/tex] = 8.00 g
Molar mass of [tex]O_2[/tex] = 32 g/mole
Molar mass of CO = 28 g/mole
First we have to calculate the moles of [tex]O_2[/tex].
[tex]\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8g}{32g/mole}=0.25moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]2C+O_2\rightarrow 2CO[/tex]
From the balanced reaction we conclude that
As, 2 mole of [tex]C[/tex] react with 1 mole of [tex]O_2[/tex]
So, 0.1 moles of [tex]C[/tex] react with [tex]\frac{0.1}{2}=0.05[/tex] moles of [tex]O_2[/tex]
From this we conclude that, [tex]O_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]C[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]CO[/tex]
From the reaction, we conclude that
As, 2 mole of [tex]C[/tex] react to give 2 mole of [tex]CO[/tex]
So, 0.1 moles of [tex]C[/tex] react to give 0.1 moles of [tex]CO[/tex]
Now we have to calculate the mass of [tex]CO[/tex]
[tex]\text{ Mass of }CO=\text{ Moles of }CO\times \text{ Molar mass of }CO[/tex]
[tex]\text{ Mass of }CO=(0.1moles)\times (28g/mole)=2.8g[/tex]
Therefore, the mass of carbon monoxide form can be 2.8 grams.
