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21 June, 00:30

If a solution containing 102.78 gof mercury (II) nitrate is allowed to react completely with a solution containing 17.796 g of sodium sulfide, how many grams of solid precipitate will form?

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  1. 21 June, 03:16
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    53 g of HgS will be formed

    Explanation:

    We determine the main reaction:

    Hg (NO₃) ₂ (aq) + Na₂S (aq) → HgS (s) ↓ + 2NaNO₃ (aq)

    We determine the limiting reactant (we need to convert the mass to moles, firstly)

    102.78 g / 324.59 g/mol = 0.316 moles of nitrate

    17.796 g / 78.06 = 0.228 moles of sulfide

    Ratio is 1:1, so the limiting reactant is the Na₂S. For 0.316 moles of nitrate, I need the same amount of sulfide and I only have 0.228 moles.

    Ratio is again 1:1. For 0.228 moles of sodium sulfide I can produce the same amount of mercury (II) sulfide

    We convert the moles to mass → 232.65 g/mol. 0.228 mol = 53 g of HgS will be formed
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