Respuesta :
Answer: 52.8 g of [tex]CO_2[/tex]
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
a) moles of octane
[tex]\text{Number of moles}=\frac{17.0g}{114g/mol}=0.150moles[/tex]
b) moles of oxygen
[tex]\text{Number of moles}=\frac{93.0g}{32g/mol}=2.91moles[/tex]
[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]
According to stoichiometry :
2 moles of [tex]C_8H_{18}[/tex] require 25 moles of [tex]O_2[/tex]
Thus 0.150 moles of [tex]C_8H_{18}[/tex] require=[tex]\frac{25}{2}\times 0.150=1.875moles[/tex] of [tex]O_2[/tex]
Thus [tex]C_8H_{18}[/tex] is the limiting reagent as it limits the formation of product and [tex]O_2[/tex] is the excess reagent.
As 2 moles of [tex]C_8H_{18}[/tex] give = 16 moles of [tex]CO_2[/tex]
Thus 0.150 moles of [tex]C_8H_{18}[/tex] give =[tex]\frac{16}{2}\times 0.150=1.2moles[/tex] of [tex]CO_2[/tex]
Mass of [tex]CO_2=moles\times {\text {Molar mass}}=1.2moles\times 44g/mol=52.8g[/tex]
Thus 52.8 g of [tex]CO_2[/tex] will be produced from the given masses of both reactants.
