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If a solution containing 84.670 g of mercury(II)chlorate .....?
cury(II) chlorate is allowed to react completely with a solution containing 12.026 g of sodium sulfide, how many grams of solid precipitate will be formed?
How many grams of the reactant in excess will remain after the reaction?
Assuming complete precipitation, how many moles of each ion remain in solution? If an ion is no longer in solution, enter a zero (0) for the number of moles.
If a solution containing 84.670 g of mercury(II)chlorate is allowed to react completely with a solution containing 12.026 g of sodium sulfide, how many grams of solid precipitate will be formed?
2 Answers
- Roger the MoleLv 74 years agoFavorite Answer
Hg(ClO3)2 + Na2S → 2 NaClO3 + HgS(s)
(84.670 g Hg(ClO3)2) / (367.4924 g Hg(ClO3)2/mol) = 0.2303993 mol Hg(ClO3)2
(12.026 g Na2S) / (78.0445 g Na2S/mol) = 0.1540916 mol Na2S
0.1540916 mole of Na2S would react completely with 0.1540916 mole of Hg(ClO3)2, but there is more Hg(ClO3)2 present than that, so Hg(ClO3)2 is in excess and Na2S is the limiting reactant.
(0.1540916 mol Na2S) x (1 mol HgS / 1 mol Na2S) x (232.655 g HgS/mol) = 35.850 g HgS
((0.2303993 mol Hg(ClO3)2 initially) - (0.1540916 mol Hg(ClO3)2 reacted)) = 0.0763077 mol Hg(ClO3)2
(0.0763077 mol Hg(ClO3)2) x (367.4924 g Hg(ClO3)2/mol) = 28.042 g Hg(ClO3)2 left over
(0.1540916 mol Na2S) x (2 mol Na{+} / 1 mol Na2S) = 0.30818 mol Na{+}
(0.0763077 mol Hg(ClO3)2) x (1 mol Hg{2+} / 1 mol Hg(ClO3)2) = (0.076308 mol Hg{2+}
(0.2303993 mol Hg(ClO3)2) x (2 mol ClO3{-} / 1 mol Hg(ClO3)2) = 0.46080 mol ClO3{-}
0 mol of S{2-}
