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What is the percent yield of ferrous sulfide if the actual yield is 220.0 g and the theoretical yield is 275.6 g?



12.5%

55.6%

79.8%

87.9%

Respuesta :

premedchemist
The percent yield is defined as:
((Actual Yield)/(Theoretical Yield)) * 100%
Plugging in our numbers we get:
(220.0 grams)/(275.6 grams) = 0.7983
Now multiply the quotient by 100%:
0.7983 * 100% = 79.83%
Closest answer is 79.8%

Answer:

79.8% is the percent yield of ferrous sulfide.

Explanation:

Percentage yield :

[tex]\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100[/tex]

Actual yield of ferrous sulfide = 220.0 g

Theoretical yield of ferrous sulfide = 275.6 g

[tex]\% (Yield)=\frac{200.0 g}{275.6 g}\times 100[/tex]

= 79.8%

79.8% is the percent yield of ferrous sulfide.

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