Respuesta :
Answer:
B) The theoretical yield of Cl2 for the reaction is 61.4505 grams
C) The percentage yield of the reaction is 93.57%
Explanations:
B) Given the reaction between HCl and Oxygen expressed as:
[tex]4HCl(g)+O_2(g)\rightarrow2H_2O(l)+2Cl_2(g)[/tex]Calculate the moles of HCl;
[tex]\begin{gathered} moles=\frac{mass}{molar\text{ mass}} \\ moles=\frac{63.1g}{36.458} \\ moles\text{ of HCl}=1.731moles \end{gathered}[/tex]Calculate the moles of Oxygen
[tex]\begin{gathered} moles\text{ of O}_2=\frac{17.2g}{32} \\ moles\text{ of O}_2=0.5375moles \end{gathered}[/tex]Since there are 4 moles of HCl, hence the limiting reactant will be HCl
According to stochiometry, you can see that 4 moles of HCl produces 2 mole of Cl2, hence the moles of Cl2 produced will be;
[tex]\begin{gathered} moles\text{ of Cl}_2=\frac{2(1.731)}{4} \\ moles\text{ of Cl}_2\text{ produced=0.8655moles} \end{gathered}[/tex]Determin the mass of Cl2 (theoretical yield)
[tex]\begin{gathered} Mass\text{ of Cl}_2=moles\times molar\text{ mass} \\ Mass\text{ of Cl}_2=0.8655\times71 \\ Mass\text{ of Cl}_2=61.4505grams \end{gathered}[/tex]Hence the theoretical yield of Cl2 for the reaction is 61.4505 grams
C) Find the percentage yield for the reaction:
[tex]\begin{gathered} \%yield=\frac{actual}{theoretical}\times100 \\ \%yield=\frac{57.5g}{61.4505}\times100 \\ \%yield=93.57\% \end{gathered}[/tex]Hence the percentage yield of the reaction is 93.57%
