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Magnesium nitrate reacts with lithium sulfate in a double replacement reaction. You have 3.45 g of lithium sulfate and 275 mL of magnesium nitrate. Calculate the molarity of the Magnesium nitrate solution in the reaction.

  • Chemistry -

    Not possible to work, as I see it. 275mL of magnesium nitrate? How can one determine how much magnesium nitrate is there? is it in excess, or deficit, or exactly enough? One wonders.

  • Chemistry -

    welp that's what the problem is.

  • Chemistry -

    First I want to make it very clear that Mg(NO3)2 and Li2SO4 will NOT react at all and it is folly to try to calculate the M of the Mg(NO3)2 this way. However, the calculation can be done (although it doesn't man anything)so in the spirit of chemistry I will continue.
    Mg(NO3)2 + Li2SO4 ==> MgSO4 + 2LiNO3

    moles Li2SO4 = grams/molar mass = ?
    Using the coefficients in the balanced equation, 1 mol Li2SO4 reacts with 1 mol Mg(NO3)2 so mol Li2SO4 = moles Mg(NO3)2
    M Mg(NO3)2 = moles Mg(NO3)2/L Mg(NO3)2

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