Question

For the following reaction, 4.41 grams of iron are mixed with excess oxygen gas . Assume that the percent yield of iron(II) oxide is 75.5 %.

iron(s) + oxygen(g)------>iron(II) oxide(s)

What is the theoretical yield of iron(II) oxide ?	grams
What is the actual yield of iron(II) oxide ?	grams

Answer 1

The balanced equation for the 2 Fe (s) + 0 (9) - 2 mole I mole 2x 56 g given reaction is. 2 FeO (s) 2 mole: = 2x72 g - 144g 32 = 112g @ from balanced equation 112 g Fe Yield = 144 g fed - 4.41 g Fe will yield = 4.41 * 144 g Feo - 112 = (5.67 g Feo Theoritical Yield 6 Since 75.5% is the yield, thus, Actual Yield = 75.8 x 5.67 100 - 4.20 g . Like if understood or comment for more