Question

CH4 (g) + 2O2(g)->CO2(g) + 2H2001)When 274.9 grams of O2 were reacted with excess CH4, 71.2 grams of CO2resulted. What was the percent yield (Hint: You'll have to calculate the theoreticalyield).

Answer 1

`\%\text{yield}=37.67\%`

Explanations:

Given the chemical reaction between methane and oxygen expressed as:

`CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)`

The formula for calculating the percent yield is given as;

`\%\text{yield}=\frac{actual\text{ yield}}{theoretical\text{ yield}}*100`

From the given information, actual yield = 71.2 grams

We can determine the theoretical yield from stochiometry.

Determine the mole of Oxygen O2.

`\begin{gathered} \text{mole of O}_2=\frac{Mass}{Molar\text{ mass}} \\ \text{mole of O}_2=(274.9g)/(2(16)) \\ \text{mole of O}_2=(274.9)/(32) \\ \text{mole of O}_2=8.59\text{moles} \end{gathered}`

According to stochiometry, 2moles of oxygen (limiting reactant) produce 1 mole of CO2, hence the number of moles of CO2 produced will be given as;

`\begin{gathered} \text{moles of CO}_2=(8.59moles)/(2) \\ \text{moles of CO}_2=4.295\text{moles} \end{gathered}`

Calculate the theoretical yield (mass of CO2)

`\begin{gathered} \text{Mass of CO}_2=moles*\text{molar mass} \\ \text{Mass of CO}_2=4.295*(12+32) \\ \text{Mass of CO}_2=4.295*44 \\ \text{Mass of CO}_2=188.994\text{grams} \end{gathered}`

Hence the theoretical yield will be 188.994 grams.

Determine the required percent yield

`\begin{gathered} \%\text{yield}=(71.2g)/(188.994)*100 \\ \%\text{yield}=0.3767*100 \\ \%\text{yield}=37.67\% \end{gathered}`