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6 July, 19:43

If a solution containing 51.429 g of mercury (ii) perchlorate is allowed to react completely with a solution containing 16.642 g of sodium sulfate, how many grams of solid precipitate will be formed?

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  1. 6 July, 20:30
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    Hg (No3) 2 + NaSO4 - - - >2NaNO3 + HgSO4 (s)

    calculate the moles of each reactant

    moles=mass/molar mass

    moles of Hg (NO3) 2 = 51.429g / 324.6 g/mol (molar mass of Hg (NO3) 2) = 0.158 moles

    moles Na2SO4 16.642g/142g/mol = 0.117 moles of Na2SO4

    Na2SO4 is the limiting reagent in the equation and by use mole ratio Na2So4 to HgSO4 is 1:1 therefore the moles of HgSO4 = 0.117 moles

    mass of HgSO4=moles x molar mass of HgSo4 = 0.117 g x 303.6g/mol = 35.5212 grams
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