Question

determine the percent yield when 56.35 g of co₂ are formed from the reaction of 4.000 moles of o₂ and excess c₈h₁₈ according to the following reaction: 2 c₈h₁₈ + 25 o₂ → 16 co₂ + 18 h₂o

Explanation:

Step1: Calculate the theoretical yield of \(CO_2\)

From the balanced chemical equation \(2C_8H_{18}+25O_2
ightarrow16CO_2 + 18H_2O\), the mole ratio of \(O_2\) to \(CO_2\) is \(25:16\).
Given \(n(O_2)=4.000\) moles.
Using the mole - ratio, \(n_{theoretical}(CO_2)=\frac{16}{25}\times n(O_2)\)
\(n_{theoretical}(CO_2)=\frac{16}{25}\times4.000\) moles \( = 2.56\) moles.
The molar mass of \(CO_2\) is \(M(CO_2)=(12.01 + 2\times16.00)\) g/mol \(=44.01\) g/mol.
The theoretical yield \(m_{theoretical}(CO_2)=n_{theoretical}(CO_2)\times M(CO_2)\)
\(m_{theoretical}(CO_2)=2.56\) moles \(\times44.01\) g/mol \(= 112.6656\) g.

Step2: Calculate the percent yield

The percent yield formula is \(\text{Percent Yield}=\frac{m_{actual}}{m_{theoretical}}\times100\%\)
Given \(m_{actual}(CO_2) = 56.35\) g and \(m_{theoretical}(CO_2)=112.6656\) g.
\(\text{Percent Yield}=\frac{56.35}{112.6656}\times100\%\)
\(\text{Percent Yield}\approx50.00\%\)

Answer:

\(50.00\%\)