Question

0.450 kg of octane burned during combustion produces 1.25 kg of carbon dioxide. what is the percent yield?

Answer 1

The percent yield is 90%

Step-by-step explanation:

First, we write the combustion reaction.

`C_(8)H_(18) + O_(2)` →
`CO_(2) + H_(2)O`

Then we balanced the equation.

2
`C_(8)H_(18) + 25O_(2)` →
`16CO_(2) + 18H_(2)O`

Now, we calculate the moles of octane.

n(octane)=
`(0,4510^(3)g )/(114,23(g)/(mol) )`

`n=3,94mol`

Now we find the theoretical moles of C02, taking into account stoichiometric coefficients

`n(CO2)=3,94mol(C8H18). (16mol(CO2))/(2mol(C8H18)) \\\\n(CO2)=31,52mol`

Finally, we calculate the mass of CO2 and proceed to calculate the percent yield

`CO_(2) (g)= 31,52mol.44(g)/(mol) \\\\CO_(2) (g)=1,39Kg`

`\\\\percent yield=(Experimental.mass)/(Theorical.mass)x100\\percent yield=(1,25Kg)/(1,39Kg)x100percent yield=89,92`

the percent yield is 89,92%≈90%