Question

Calculate percent yield of 1.4736g K(Fe(C2O4)2)·4H2O OBTAINED FROM starting compound 1.2770g (NH4)2Fe(SO4)2·6H2O again 1.2770g = starting...

Calculate percent yield of 1.4736g K(Fe(C2O4)2)·4H2O OBTAINED FROM starting compound 1.2770g (NH4)2Fe(SO4)2·6H2O

again 1.2770g = starting material

1.4736g = product

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Answer #1

The balanced equation is :

2Fe(NH4)2(SO4)2*6H2O +3H2C2O4 + 3 K2C2O4 + H2O2 -> 2K3Fe(C2O4)3*3H2O + 2(NH4)2SO4 + 2H2SO4 + 6H2O

392 491

2 moles = 2*392 g of Fe(NH4)2(SO4)2*6H2O produces 2 moles = 2x491 g of K3Fe(C2O4)3*3H2O

1.2770g of Fe(NH4)2(SO4)2*6H2O produces (1.2770*2x491)/(2*392)= 1.5995 g of K3Fe(C2O4)3*3H2O

So theoretical yield is 1.5995 g

Given actual yield is 1.4736 g

Percent yield is = (actual yield/theoretical yield ) x 100

= (1.4736/1.5995)x100

= 92.1 %

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