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the synthesis of FeC2O4*2H2O, Which reagent would be the limiting reagent? (NH4)2[Fe(H2O)6](SO4)2 IS 3.42g H2C2O4*2H2O is...

the synthesis of FeC2O4*2H2O, Which reagent would be the limiting reagent? (NH4)2[Fe(H2O)6](SO4)2 IS 3.42g H2C2O4*2H2O is 1.98g 2. using your answers above, calculate the theoretical yield (in grams) of potassium trioxalatofer-rate(iii) trihydrate, K3[Fe(C2O4)3]*3H2O

Solutions

Expert Solution

1. First write down the balanced equation for the preparation of FeC2O4.2H2O

The balanced equation is:

Fe(NH4)2(SO4)2.6H2O(s) + H2C2O4(aq) FeC2O4.2H2O(s)+ (NH4)2SO4(aq) + H2SO4(aq) + 4H2O

Ferrous Ammonium oxalic acid Ferrous oxalate ammonium

sulphate hexahydrate dihydrate sulfate

mass of Fe(NH4)2(SO4)2.6H2O = 3.42 g

mass of H2C2O4.2H2O = 1.98 g

Calculate the moles of Fe(NH4)2(SO4)2.6H2O = 3.42 g * 1 mol Fe(NH4)2(SO4)2.6H2O / 392.14 g = 0.00872 mol

Now calculate the moles of H2C2O4.2H2O = 1.98 g * 1mol H2C2O4.2H2O / 126.07 g = 0.0157 mol

There are less moles of Fe(NH4)2(SO4)2.6H2O , so it is a limiting reagent.

According to the balanced equation, there is 1:1 molar ratio between Fe(NH4)2(SO4)2.6H2O and FeC2O4.2H2O.

Hence, Theoretical yield of FeC2O4.2H2O = 0.00872 mol

2. The balanced equation for the preparation of K3Fe(C2O4)3.3H2O from FeC2O4.2H2O

2FeC2O4.2H2O(s) + H2C2O4(aq) + H2O2(aq)+3K2C2O4(aq) 2 K3[Fe(C2O4)3].3H2O(s)

There is 1:1 molar ratio between FeC2O4.2H2O and K3[Fe(C2O4)3].3H2O

Hence, Theoretical yield of K3Fe(C2O4)3.3H2O (considering FeC2O4.2H2O as limiting reagent) = 0.00872 mol

Theoretical yield of K3Fe(C2O4)3.3H2O in grams = 0.00872 mol * 491.23 g K3Fe(C2O4)3.3H2O / 1 mol = 4.28 g

queen_honey_blossom answered 2 years ago
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