Respuesta :
Answer: The percent yield is, 42%
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of carbon disulphide}=\frac{1.55g}{76g/mol}=0.020moles[/tex]
[tex]\text{Moles of oxygen}=\frac{2.83g}{32g/mol}=0.088moles[/tex]
The balanced chemical reaction is:
[tex]CS_2+3O_2(g)\rightarrow CO_2+2SO_2[/tex]
According to stoichiometry :
1 moles of [tex]CS_2[/tex] require = 3 moles of [tex]O_2[/tex]
Thus 0.020 moles of [tex]CS_2[/tex] will require=[tex]\frac{3}{1}\times 0.020=0.060moles[/tex] of [tex]O_2[/tex]
Thus [tex]CS_2[/tex] is the limiting reagent as it limits the formation of product and [tex]O_2[/tex] is the excess reagent.
As 1 mole of [tex]CS_2[/tex] give = 2 moles of [tex]SO_2[/tex]
Thus 0.020 moles of [tex]CS_2[/tex] give =[tex]\frac{2}{1}\times 0.020=0.040moles[/tex] of [tex]SO_2[/tex]
Theoretical mass of [tex]SO_2=moles\times {\text {Molar mass}}=0.040moles\times 64g/mol=2.56g[/tex]
Actual mass of [tex]SO_2[/tex] = 1.1 g
Now we have to calculate the percent yield
[tex]\%\text{ yield}=\frac{\text{Actual yield }}{\text{Theoretical yield}}\times 100=\frac{1.1g}{2.56g}\times 100=42\%[/tex]
Therefore, the percent yield is, 42%
