What is the percent yield of ferrous sulfide if the actual yield is 220.0 g and the theoretical yield is 275.6 g?
Answer (3)
If the actual yield of ferrous sulfide is 220.0 g and its theoretical yield is 275.6 g, the percent yield of ferrous sulfide is 79.8%. Make a proportion. 220.0 g is x percent. 275.6 g is 100%. 220.0 g : x = 275.6 g : 100%. After crossing the products: x = 220.0 g * 100% : 275.6 g = 79.8%
To get the percent yield, we will use this formula: ((Actual Yield)/(Theoretical Yield)) * 100% Values given: actual yield is 220.0 g theoretical yield is 275.6 g Now, let us substitute the values given. (220.0 grams)/(275.6 grams) = 0.7983 Then, to get the percentage, multiply the quotient by 100. 0.7983 (100) = 79.83% Among the choices, the most plausible answer is 79.8%
The percent yield of ferrous sulfide is approximately 79.8%, calculated by dividing the actual yield of 220.0 g by the theoretical yield of 275.6 g and multiplying by 100. This percentage indicates the efficiency of the chemical reaction. Understanding percent yield is important for evaluating the success of chemical processes.
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