Chemistry Help Needed

Question

The first step of the synthesis is described by the reaction below. When 1.500 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)26H2O(s) H2C2O4(aq)

FeC2O42H2O(s) (NH4)2SO4(aq) H2SO4(aq) 4 H2O(l)

Answer 1

Question

Chemistry Help Needed

The first step of the synthesis is described by the reaction below. When 1.500 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 dot 6H2O(s)  & H2C2O4(aq) --> FeC2O42H2O(s)

let's find moles:

1.500 grams Fe(NH4)2(SO4)2 dot 6H2O @ 392.14 g/mol = 3.825e-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 1.500 grams Fe(NH4)2(SO4)2 dot 6H2O is the limiting reagent. so how many grams of FeC2O4dot2H2O can be produced  form the 1.500 grams Fe(NH4)2(SO4)2 .... using the molar masses of each:

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

0.6882 grams of FeC2O4dot2H2O

Your answer (4 sig figs) : 0.6882 grams of FeC2O4dot2H2O