Respuesta :
Answer: The percent yield in this experiment is 63.1 %
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} CaO=\frac{1.00g}{56g/mol}=0.018moles[/tex]
[tex]CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(s)[/tex]
As [tex]H_2O[/tex] is the excess reagent, [tex]CaO[/tex] acts as the limiting reagent and it limits the formation of product.
According to stoichiometry :
1 mole of [tex]CaO[/tex] produce = 1 mole of [tex]Ca(OH)_2[/tex]
Thus 0.018 moles of [tex]CaO[/tex] will produce=[tex]\frac{1}{1}\times 0.018=0.018moles[/tex] of [tex]Ca(OH)_2[/tex]
Mass of [tex]Ca(OH)_2=moles\times {\text {Molar mass}}=0.018moles\times 74g/mol=1.3g[/tex]
Theoretical yield = 1.3 g
Experimental yield of ethanol = 0.82 g
[tex]\%\text{ yield }=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100=\frac{0.82g}{1.3g}\times 100=63.1\%[/tex]
Thus the percent yield in this experiment is 63.1 %
