# A 10*g mass of carbon dioxide is obtained from the combustion of a 70.2*g mass of octane. What is the apparent yield?

May 2, 2017

We interrogate the stoichiometric combustion equation..............and the yield is under 5%.

#### Explanation:

${C}_{8} {H}_{18} \left(l\right) + \frac{25}{2} {O}_{2} \left(g\right) \rightarrow 8 C {O}_{2} \left(g\right) + 9 {H}_{2} O \left(l\right)$

$\text{Moles of octane} = \frac{70.2 \cdot g}{114.23 \cdot g \cdot m o {l}^{-} 1} = 0.614 \cdot m o l$

"Yield of carbon dioxide"=(10.0*gxx100%)/(8xx0.614*molxx44.01*g*mol^-1)=??%

Are sure you've copied down the appropriate data? The yield seems a little low.