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Oxalic acid (H2C2O4) can be used to remove rust according to the following balanced equation:

Fe2O3 + 6H2C2O4 --> 2Fe(C2O4)33- +3H2O + 6H+

Calculate the number of grams of rust that can be removed by 5.00 x 102 ml of a 0.122M solution of oxalic acid.

I know i need to multiply the # of moles*6*160 (which is the molecular mass)
0.061 moles of oxalic acid.
so is it .061*6*160= 585.6
is that right i'm pretty sure its not

  • chemistry -

    The oxalic acid solution contains
    0.500 x 0.122 moles of oxalic acid.

    As 6 moles of oxalic acid react with 1 moles of rust then 0.500 x 0.122 moles will react with
    0.500 x 0.122 x 1/6 moles of rust

    the MM for rust is 159.6882 g mole^-1

    so the mass of rust is

    0.500 x 0.122 x 1/6 x 159.6882 g

    (3 sig figs are appropriate int he answer)

  • chemistry -

    Your answer leaves me with 1.63 so is this wrong moles oxalic acid = 0.500 L x 0.122 M=0.610

    the ratio between Fe2O3 and H2C2O4 is 1 : 6

    moles Fe2O3 = 0.610/6=0.102

    mass Fe2O3 = 0.102 mol x 159.69 g/mol=16.3 g
    It left me with 16.3

  • chemistry -

    You error is here

    0.500 L x 0.122 M = 0.061 moles and NOT 0.610 moles

    I got 1.63 g agreed

  • chemistry -

    oh okay i got it

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