Respuesta :
Answer : The theoretical yield of [tex]CO_2[/tex] is, 112.64 grams.
Explanation : Given,
Given moles of [tex]C_8H_{18}[/tex] = 4 moles
Given moles of [tex]O_2[/tex] = 4 moles
Molar mass of [tex]CO_2[/tex] = 44 g/mole
First we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]
From the given balanced reaction, we conclude that
As, 25 moles of [tex]O_2[/tex] react with 2 moles of [tex]C_8H_{18}[/tex]
So, 4 moles of [tex]O_2[/tex] react with [tex]\frac{2}{25}\times 4=0.32[/tex] moles of [tex]C_8H_{18}[/tex]
From this we conclude that, [tex]C_8H_{18}[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]O_2[/tex] is a limiting reagent because it limits the formation of product.
Now we have to calculate the moles of [tex]CO_2[/tex].
As, 25 moles of [tex]O_2[/tex] react to give 16 moles of [tex]CO_2[/tex]
So, 4 moles of [tex]O_2[/tex] react to give [tex]\frac{16}{25}\times 4=2.56[/tex] moles of [tex]CO_2[/tex]
Now we have to calculate the mass of [tex]CO_2[/tex].
[tex]\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2[/tex]
[tex]\text{Mass of }CO_2=(2.56mole)\times (44g/mole)=112.64g[/tex]
Therefore, the theoretical yield of [tex]CO_2[/tex] is, 112.64 grams.
