Question

c. Given an actual yield of 0.99 g iron(II) oxalate, FeC204.2H20, calculate the percentage yield

Answer 1

The balanced equation is 2 Fe(NH4)2(SO4)2 6 H2O (aq) + 2 H2C204 --------> 2 FeC204. 2 H2O + (NH4)2(SO4) + 2 H2SO4 + 4H2O Number of moles of Fe(NH4)2(SO4)2.6H2O = mass/mol.wt. = 2.5 g/(392.14 g/mol) = 0.006375 mol Number of moles of H2C204 = 1.5 M * 10.00 mL = 15 mmol = 0.015 mol According to balanced equation, 2 moles of Fe(NH4)2(SO4)2.6H2O reacts with 2 mole of H2C204 So, 0.006375 mol of Fe(NH4)2(SO4)2.6H,O would react with 0.006375 mol of H, C,04 But, the number of moles of H,C,04 present in excess That is H2C204 is the excess reagent Fe(NH)2(SO4)2.6H2O is the Limiting reagent. (b) According to balanced equation, 2 moles of Fe(NH4)2(SO4)2.6H20 produces 2 mole of FeC204. 2 H2O So, 0.006375 mol of Fe(NH4)2(SO4)2.6H2O would produce 0.006375 mol of FeC204.2 H2O Mass of FeC,04.2 HO = moles * mol.wt. = 0.006375 mol * 179.89 g/mol = 1.146848 g Theoretical Yield of FeC204. 2 H20 = 1.147 g (c) Percent Yield = Actual Yield /Theoretical yield *100 Percent Yield = 0.99 / 1.147 *100 = 86.32 %