Question

give the percent yield when 28.16 g of co2 are formed from the reaction of 4.000 moles of c8h18 with 4.000 moles of o2.

2 c8h18 + 25 o2 → 16 co2 + 18 h2o

Explanation:

Step1: Determine the limiting reactant

The mole - ratio of $\ce{C8H18}$ to $\ce{O2}$ from the balanced equation $2\ce{C8H18}+25\ce{O2}
ightarrow16\ce{CO2}+18\ce{H2O}$ is $\frac{n_{\ce{C8H18}}}{n_{\ce{O2}}}=\frac{2}{25}= 0.08$. Given $n_{\ce{C8H18}} = 4.000$ mol and $n_{\ce{O2}}=4.000$ mol, the mole - ratio of the reactants is $\frac{4.000}{4.000}=1$. Since $1>0.08$, $\ce{O2}$ is the limiting reactant.

Step2: Calculate the theoretical yield of $\ce{CO2}$

From the balanced equation, the mole - ratio of $\ce{O2}$ to $\ce{CO2}$ is $\frac{n_{\ce{CO2}}}{n_{\ce{O2}}}=\frac{16}{25}$. Given $n_{\ce{O2}} = 4.000$ mol, then $n_{\ce{CO2}_{theo}}=\frac{16}{25}\times4.000$ mol$ = 2.56$ mol. The molar mass of $\ce{CO2}$ is $M = 44.01$ g/mol. So the theoretical mass of $\ce{CO2}$, $m_{\ce{CO2}_{theo}}=n_{\ce{CO2}_{theo}}\times M=2.56$ mol$\times44.01$ g/mol$=112.6656$ g.

Step3: Calculate the percent yield

The percent - yield formula is $\text{Percent Yield}=\frac{m_{\ce{CO2}_{actual}}}{m_{\ce{CO2}_{theo}}}\times100\%$. Given $m_{\ce{CO2}_{actual}} = 28.16$ g and $m_{\ce{CO2}_{theo}}=112.6656$ g. Then $\text{Percent Yield}=\frac{28.16}{112.6656}\times100\%\approx25.00\%$.

Answer:

$25.00\%$