Answer to Question #115346 in General Chemistry for Jailene

Question

If, in the reaction below, 83.7 grams of Fe producers 150.0 grams of Fe2S3 what is the % yield?

Answer 1

83.7g________________________x

2Fe____=>_____Fe₂S₃

2*55.85 g/mol_________(2*55.85+3*32.06)g/mol = 207.88 g/mol

In case of 100% yield reaction produces:

x = m (Fe₂S₃) = 83.7 * 207.88 / (2*55.85) = 155.77 g, so the yield is:

w = (150/155.77) * 100% = 96.3%