Respuesta :
Answer:
The percent yield is 90%
Explanation:
First, we write the combustion reaction.
[tex]C_{8}H_{18} + O_{2}[/tex] → [tex]CO_{2} + H_{2}O[/tex]
Then we balanced the equation.
2[tex]C_{8}H_{18} + 25O_{2}[/tex] → [tex]16CO_{2} + 18H_{2}O[/tex]
Now, we calculate the moles of octane.
n(octane)=[tex]\frac{0,4510^{3}g }{114,23\frac{g}{mol} }[/tex]
[tex]n=3,94mol[/tex]
Now we find the theoretical moles of C02, taking into account stoichiometric coefficients
[tex]n(CO2)=3,94mol(C8H18). \frac{16mol(CO2)}{2mol(C8H18)} \\\\n(CO2)=31,52mol[/tex]
Finally, we calculate the mass of CO2 and proceed to calculate the percent yield
[tex]CO_{2} (g)= 31,52mol.44\frac{g}{mol} \\\\CO_{2} (g)=1,39Kg[/tex]
[tex]\\\\percent yield=\frac{Experimental.mass}{Theorical.mass}x100\\percent yield=\frac{1,25Kg}{1,39Kg}x100percent yield=89,92[/tex]
the percent yield is 89,92%≈90%
