Respuesta :
Answer:
[tex]\%\text{yield}=37.67\%[/tex]Explanations:
Given the chemical reaction between methane and oxygen expressed as:
[tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)[/tex]The formula for calculating the percent yield is given as;
[tex]\%\text{yield}=\frac{actual\text{ yield}}{theoretical\text{ yield}}\times100[/tex]From the given information, actual yield = 71.2 grams
We can determine the theoretical yield from stochiometry.
Determine the mole of Oxygen O2.
[tex]\begin{gathered} \text{mole of O}_2=\frac{Mass}{Molar\text{ mass}} \\ \text{mole of O}_2=\frac{274.9g}{2(16)} \\ \text{mole of O}_2=\frac{274.9}{32} \\ \text{mole of O}_2=8.59\text{moles} \end{gathered}[/tex]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;
[tex]\begin{gathered} \text{moles of CO}_2=\frac{8.59moles}{2} \\ \text{moles of CO}_2=4.295\text{moles} \end{gathered}[/tex]Calculate the theoretical yield (mass of CO2)
[tex]\begin{gathered} \text{Mass of CO}_2=moles\times\text{molar mass} \\ \text{Mass of CO}_2=4.295\times(12+32) \\ \text{Mass of CO}_2=4.295\times44 \\ \text{Mass of CO}_2=188.994\text{grams} \end{gathered}[/tex]Hence the theoretical yield will be 188.994 grams.
Determine the required percent yield
[tex]\begin{gathered} \%\text{yield}=\frac{71.2g}{188.994}\times100 \\ \%\text{yield}=0.3767\times100 \\ \%\text{yield}=37.67\% \end{gathered}[/tex]