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    A sample (1.1534g) of a tin containing substance was converted into an aqueous Sn2+ solution with a total volume of 250 cm^3. Three 25 cm^3 aliquots of this Sn2+ solution were titrated with a KMnO4 solution of concentration 1.78 x 10^-2M. The titres were recorded as 16.53 + 16.49 + 16.56 cm^3.

    Calculate the total number of moles of Sn2+ and the concentration of Sn2+ in the original solution.


    So i have the equation:
    5Sn2+ + 2MnO4- + 16H+ --> 5Sn4+ + 2Mn2+ + 8H2O


    Number of moles:

    Take the average of the 3 titre volumes: 0.01653 dm^3
    1.78x10^-2 * 0.01653 = 2.94 x 10^-4
    Divide by 2 and then multiply by 5: 7.35 x 10^-4 moles

    Concentration:

    (7.35 x 10^-4) / 0.01653 = 0.044 mol dm^-3

    Is this at all right, or have I done this completely wrong?

    Also, how would I calculate the total mass of Sn and percentage by mass of Sn in the original sample?
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    (Original post by YesNoMaybeIDontKnow)
    A sample (1.1534g) of a tin containing substance was converted into an aqueous Sn2+ solution with a total volume of 250 cm^3. Three 25 cm^3 aliquots of this Sn2+ solution were titrated with a KMnO4 solution of concentration 1.78 x 10^-2M. The titres were recorded as 16.53 + 16.49 + 16.56 cm^3.

    Calculate the total number of moles of Sn2+ and the concentration of Sn2+ in the original solution.


    So i have the equation:
    5Sn2+ + 2MnO4- + 16H+ --> 5Sn4+ + 2Mn2+ + 8H2O


    Number of moles:

    Take the average of the 3 titre volumes: 0.01653 dm^3
    1.78x10^-2 * 0.01653 = 2.94 x 10^-4
    Divide by 2 and then multiply by 5: 7.35 x 10^-4 moles

    Concentration:

    (7.35 x 10^-4) / 0.01653 = 0.044 mol dm^-3
    Is this at all right, or have I done this completely wrong?

    Also, how would I calculate the total mass of Sn and percentage by mass of Sn in the original sample?
    You have gone a little astray (red)

    Heres a run through of the problem clickme (turn speakers on!)

    The final part about the percentage is simply to multiply the moles of Sn by the relative mass and then see what percentage of the original masss it comes to... (I'll leave that to you)
 
 
 
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