Question

Consider the reaction: 2Al + 3S = Al2S3If you have 45 grams of aluminum...a. How many grams of aluminum sulfide are produced? (Theoretical)b. When the rxn is performed in the lab, 114 grams of aluminum sulfide is produced. What is the percent yield?

Answer 1

The reaction they give us is balanced. We must determine the moles of aluminum that are equivalent to that 45 g, for this, we will use the molar mass of aluminum equal to 26.98g/mol

`\begin{gathered} \text{Mol of Al}=GivengAl*(1molAl)/(MolarMass,gAl) \\ \text{Mol of Al}=45gAl*(1molAl)/(26.98gAl)=1.67molAl \end{gathered}`

Now, in the reaction, we can see that for every 2 moles of aluminum, 1 mole of aluminum sulfate is produced. We can also use the molar mass of aluminum sulfate (101.96g/mol) to calculate the grams of aluminum sulfide produced.

`\begin{gathered} gAl_2O_3=GivenMolAl*(1molAl_2O_3)/(2molAl)*(MolarMass,gAl_2O_3)/(1molAl_2O_3) \\ gAl_2O_3=1.67molAl*(1molAl_2O_3)/(2molAl)*(101.96gAl_2O_3)/(1molAl_2O_3)=85gAl_2O_3 \end{gathered}`

The grams of aluminum sulfate produced is 85g

The percent yield is calculated with the next equation:

`\begin{gathered} \text{Percent yield=}\frac{\text{Actual yield}}{Theoretical\text{ yield}}*100\% \\ \text{Percent yield=}\frac{\text{1}14g}{85g}*100\%=134\% \end{gathered}`

Percent yield=134%

This is not experimentally possible