If 25.0 mL of 0.100 M lithium iodide reacts completely with aqueous mercury(II)nitrate,what is the mass of HgI₂ (454.39 g/mol)precipitate? 2 LiI(aq)+ Hg(NO₃)2(aq)→ HgI₂(s)+ 2 LiNO₃(aq)

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Quizplus · Accepted Answer

The moles of LiI are 0.0025 mol (25.0 mL * 0.100 M). The reaction shows a 1:1 molar ratio between LiI and HgI₂, so 0.0025 mol of HgI₂ is formed. The mass of HgI₂ is 0.0025 mol * 454.39 g/mol = 1.14 g.

If 250 ML of 0