Question

This requires you to use all the things we have learned in this chapter. 1.55 grams of Carbon disulfide

Answer 1

Answer: The percent yield is, 42%

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of carbon disulphide}=\frac{1.55g}{76g/mol}=0.020moles$

$\text{Moles of oxygen}=\frac{2.83g}{32g/mol}=0.088moles$

The balanced chemical reaction is:

$CS_2+3O_2(g)\rightarrow CO_2+2SO_2$

According to stoichiometry :

1 moles of $CS_2$ require = 3 moles of $O_2$

Thus 0.020 moles of $CS_2$ will require=$\frac{3}{1}\times 0.020=0.060moles$  of $O_2$

Thus $CS_2$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

As 1 mole of $CS_2$ give = 2 moles of $SO_2$

Thus 0.020 moles of $CS_2$ give =$\frac{2}{1}\times 0.020=0.040moles$  of $SO_2$

Theoretical mass of $SO_2=moles\times {\text {Molar mass}}=0.040moles\times 64g/mol=2.56g$

Actual mass of $SO_2$ = 1.1 g

Now we have to calculate the percent yield

$\%\text{ yield}=\frac{\text{Actual yield }}{\text{Theoretical yield}}\times 100=\frac{1.1g}{2.56g}\times 100=42\%$

Therefore, the percent yield is, 42%