Finding the limiting reactant in the formation of Fe(C_2O_4) middot 2H_2O. Calcu

Question

Finding the limiting reactant in the formation of Fe(C_2O_4) middot 2H_2O. Calculate the number of moles of ferrous ammonium sulfate hexahydrate that you used. Calculate the number of moles of oxalic acid that was used in the formation of the yellow salt. What is the ratio or ferrous ion to oxalate ion? And which is the limiting reactant? Finding the ratio of ferrous to oxalate used in the formation of the green salt. Calculate the number of moles of potassium oxalate used. Calculate the number of moles of additional oxalic acid used. Add all of the moles of oxalate ion together (3b, 4a, and 4b). What is the ratio of ferrous ion to oxalate ion in the formation of the green salt? Given that the hydrogen peroxide solution is 3% (meaning 3 grams of hydrogen peroxide in 100 ml. of solution), what is the concentration in molarity? Calculate the number of moles of hydrogen peroxide used in the synthesis of the green salt. Which is the limiting reactant, ferrous ion or hydrogen peroxide? You will need to use your calculations from question 3 and 6.

Explanation / Answer

Data not given for the mass of ferrous ammonium sulfate hexahydrate(NH4)2Fe(SO4)2.6H2O

oxalic acid =(COOH)2

(NH4)2Fe(SO4)2.6H2O(s)+ (COOH)2(aq) --->FeC2O4.2H2O(s)+(NH4)2SO4(aq) +H2SO4(aq) +4H2O(l)

1) Let the mass of ferrous ammonium sulfate hexahydrate be x grams

moles of ferrous ammonium sulfate hexahydrate =x g/molar mass of ferrous ammonium sulfate hexahydrate=x/392.125 g/mol=x/392.125 moles

2)From the balanced equation ,moles of ferrous ammonium sulfate hexahydrate reacted with equal moles of oxalic acid(see the coefficients of both reactant is 1 in the balanced equation)

So moles of oxalic acid used up =x/392.125 moles

3)limiting reactant is the reactant which is present in lesser amount .

compare the moles of ferrous ammonium sulfate hexahydrate and moles of oxalic acid .

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