For the following reaction, 6.30 grams of oxygen gas are mixed with excess iron.

Question

For the following reaction, 6.30 grams of oxygen gas are mixed with excess iron. The reaction yields 18.9 grams of iron(II) oxide.

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

What is the theoretical yield of iron(II) oxide ? ___?___grams
What is the percent yield for this reaction ? ___?___%

Explanation / Answer

2 Fe + O2 ======> 2FeO Moles = mass/molar mass Moles of O2 = 6.30 g/32 g/mol = 0.20 mol O2 Here the molar ratio is 1 O2 : 2 FeO 0.20 mol O2 * (2 mol FeO/ 1mol O2)*(71.84 g FeO/1 mol FeO) 28.74 g FeO [Theoretical] % of yield = actual /theroretical * 100 % of yield = (18.9/28.74)*100 = 65.76 %

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