What is the theoretical yield of...

Question

The first step of the synthesis is described by the reaction below. When 1.200 g of Fe(NH4)2(SO4)2 *6H2O is mixed with 13 mL of 1.0 M H2C2O4, the theoretical yield of FeC2O42H2O is__________grams.

Fe(NH4)2(SO4)2*6H2O(s) H2C2O4(aq)

FeC2O4 2H2O(s) (NH4)2SO4(aq) H2SO4(aq) 4 H2O(l)

Answer 1

Fe(NH4)2(SO4)2 dot 6H2O(s) & H2C2O4(aq) --> FeC2O42H2O(s)

let's find moles:

1.200 grams Fe(NH4)2(SO4)2 dot 6H2O @ 392.14 g/mol = 3.060e-3 moles

0.013 litres @ 1 mol/ litre = 0.013 moles H2C2O4

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since they react 1 mole to 1 mole , the 1.200 grams Fe(NH4)2(SO4)2 dot 6H2O is the limiting reagent. so how many grams of FeC2O4dot2H2O can be produced form the 1.200 grams Fe(NH4)2(SO4)2 .... using the molar masses of each:

1.200 g @ 179.92 g/mol FeC2O4dot2H2O / 392.14 g/mol Fe(NH4)2(SO4)2 dot 6H2O =

0.5506 grams of FeC2O4dot2H2O

Your answer (4 sig figs) : 0.5506 grams of FeC2O4dot2H2O