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Using the balanced equation just found and the mass of fe(nh4)2(so4)2·6h2o recorded at the beginning of last week's experiment determine the percent yield of kxfey(c2o4)z.nh2o .

Respuesta :

meerkat18
The balanced equation is Fe(NH4)2(SO4)2·6H2O + H2C2O4 + 2K2C2O4 + H2O2 →K4Fe1(C2O4)3.nH2O + (NH4)2SO4 + H2SO4 + 8H2O

Percent Yield is actual/theoretical x 100

Actual is the 2.068g we measured after vacuum filtering in part c

Theoretical is 

2.522g Fe(NH4)2(SO4)2·6H2O x (1 mol Fe(NH4)2(SO4)2·6H2O/ 392.14g) x (1 mol K4Fe(C2O4)3 nH2O/1mol Fe(NH4)2(SO4)2·6H2O) x (346.1102 gK4Fe(C2O4)3 nH2O/ 1 mol K4Fe(C2O4)3 nH2O) = 2.226g

Therefore percent yield is (2.068g/2.226g) x 100= 92.9%

The percent yield of the compound from the balanced chemical reaction has been 92.9%.

The balanced equation for the reaction will be:

[tex]\rm Fe(NH_4)_2SO_4_2.6H_2O\;+\;K_2C_2O_4\;+\;H_2C_2O_4\;+\;H_2O_2\;\rightarrow\;K_4Fe(C_2O_4)_3.nH_2O\;+\;(NH_4)_2SO_4\;+\;H_2SO_4\;+\;8\;H_2O[/tex]

The percent yield can be given by:

Percent yield = [tex]\rm \dfrac{Actual\;yield}{Theoretical\;yield}\;\times\;100[/tex]

The actual yield has been = 2.226 grams

The theoretical yield has been = 20.68 grams

Percent yield = [tex]\rm \dfrac{2.068}{2.226}\;\times\;100[/tex]

Percent yield = 92.9%

The percent yield of the compound from the balanced chemical reaction has been 92.9%.

For more information about the  percent yield, refer to the link:

https://brainly.com/question/2506978

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