Question

Consider the reaction between HCl and O2:4 HCI (9) + 02 (9) 2 H20 (1) +2 C12 (9)When 63.1 g of HCA are allowed to react with 17.2g of O2, 57.5 g of Cl, are collected.Part BDetermine the theoretical yield of Cl2 for the reaction.m=gPart CDetermine the percent yield for the reaction.percent yield%

Answer 1

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:

`4HCl(g)+O_2(g)\rightarrow2H_2O(l)+2Cl_2(g)`

Calculate the moles of HCl;

`\begin{gathered} moles=\frac{mass}{molar\text{ mass}} \\ moles=(63.1g)/(36.458) \\ moles\text{ of HCl}=1.731moles \end{gathered}`

Calculate the moles of Oxygen

`\begin{gathered} moles\text{ of O}_2=(17.2g)/(32) \\ moles\text{ of O}_2=0.5375moles \end{gathered}`

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;

`\begin{gathered} moles\text{ of Cl}_2=(2(1.731))/(4) \\ moles\text{ of Cl}_2\text{ produced=0.8655moles} \end{gathered}`

Determin the mass of Cl2 (theoretical yield)

`\begin{gathered} Mass\text{ of Cl}_2=moles* molar\text{ mass} \\ Mass\text{ of Cl}_2=0.8655*71 \\ Mass\text{ of Cl}_2=61.4505grams \end{gathered}`

Hence the theoretical yield of Cl2 for the reaction is 61.4505 grams

C) Find the percentage yield for the reaction:

`\begin{gathered} \%yield=(actual)/(theoretical)*100 \\ \%yield=(57.5g)/(61.4505)*100 \\ \%yield=93.57\% \end{gathered}`

Hence the percentage yield of the reaction is 93.57%