Respuesta :
This is an incomplete question, here is a complete question.
Calculate the percent yield when 2 mol of metal antimony (atomic mass 121.76 amu) and 5 mol of chlorine (molar mass 70.90) to form 444.6 g of antimony trichloride (molar mass 228.0 amu).
The chemical reaction is the following:
[tex]4Sb+6Cl_2\rightarrow 4SbCl_3[/tex]
Answer : The percent yield of the reaction is, 9.75 %
Explanation :
First we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]4Sb+6Cl_2\rightarrow 4SbCl_3[/tex]
From the balanced reaction we conclude that
As, 4 mole of [tex]Sb[/tex] react with 6 mole of [tex]Cl_2[/tex]
So, 2 moles of [tex]Sb[/tex] react with [tex]\frac{6}{4}\times 2=3[/tex] moles of [tex]Cl_2[/tex]
From this we conclude that, [tex]Cl_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Sb[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]SbCl_3[/tex]
From the reaction, we conclude that
As, 4 mole of [tex]Sb[/tex] react to give 4 mole of [tex]SbCl_3[/tex]
So, 2 mole of [tex]Sb[/tex] react to give 2 mole of [tex]SbCl_3[/tex]
Now we have to calculate the mass of [tex]SbCl_3[/tex]
[tex]\text{ Mass of }SbCl_3=\text{ Moles of }SbCl_3\times \text{ Molar mass of }SbCl_3[/tex]
Molar mass of [tex]SbCl_3[/tex] = 228.0 g/mole
[tex]\text{ Mass of }SbCl_3=(2moles)\times (228.0g/mole)=456.0g[/tex]
Now we have to calculate the percent yield of the reaction.
[tex]\text{Percent yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Percent yield}=\frac{444.6g}{456.0g}\times 100=9.75\%[/tex]
Therefore, the percent yield of the reaction is, 9.75 %
