If a solution containing 84.670 g of mercury(II) nitrate is allowed to react completely with a...

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If a solution containing 84.670 g of mercury(II) nitrate isallowed to react completely with a solution containing 12.026 g ofsodium sulfide, how many grams of solid precipitate will beformed?How many grams of the reactant in excess will remain afterthe reaction?

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4.2 Ratings (718 Votes)

Hg(NO3)2 +Â Â Na2S ----------------> HgS (s) + 2 NaNO3

324.6gÂ Â Â Â Â Â Â Â Â Â Â 78.04gÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 232.65g

84.670gÂ Â Â Â Â Â Â Â Â Â Â 12.026g

here limiting reagent is Na2S . so the product based on that

mass of precipitate HgS formed = 12.026 x 232.65 / 78.04

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 35.85 g

mass of precipitate HgS formed Â Â = 35.85 g

mercury(II) nitrate consumed = 50.02

mercury(II) nitrate excess will remain after the reaction = 84.670 - 50.02

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 34.649 g

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