Find the limiting reactant in the formation of Fe(C2O4).2H2O = 7.368g?

35mL of a 1 M solution of oxalic acid was added

Update:

7.368 g is the mass of Fe(NH4)2(SO4)2.6H20

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  • 9 months ago

    Limiting reactant problem ....

    Fe(NH4)2(SO4)2•6H2O + H2C2O4 --> Fe(C2O4)•2H2O + 2(NH4)2SO4

    ............. 7.368 g ............... 35.0 mL ...............?g

    .................. ....................... 1.00M

    7.268g Fe(NH4)2(SO4)2•6H2O x (1 mol Fe(NH4)2(SO4)2•6H2O / 394.2 g Fe(NH4)2(SO4)2•6H2O) = 0.01844 mol Fe(NH4)2(SO4)2•6H2O

    0.0350L x (1 mol H2C2O4 / 1L) = 0.0350 mol H2C2O4

    Since the mole ratio is 1:1, Fe(NH4)2(SO4)2•6H2O is the limiting reactant, and 0.01844 mol Fe(C2O4)•2H2O is produced.

    And should you need to know the mass of the product....

    0.01844 mol Fe(C2O4)•2H2O x (179.89g Fe(C2O4)•2H2O / 1 mol Fe(C2O4)•2H2O = 3.317g Fe(C2O4)•2H2O

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