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enot [183]
4 years ago
10

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

Biology
1 answer:
Alja [10]4 years ago
4 0

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%

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