Final answer:
The mass of solid precipitate (mercury (II) sulfide) formed when mercury (II) chlorate reacts with sodium sulfide is 39.37 grams.
Step-by-step explanation:
In the given reaction,
Mercury (II) Chlorate (Hg(ClO3)2) reacts with sodium sulfide (Na2S) to form a solid precipitate of mercury (II) sulfide (HgS):
Hg(ClO3)2 + Na2S → HgS + 2NaClO3
To determine the amount of solid precipitate formed, we need to calculate the molar masses of Hg(ClO3)2 and Na2S, and then use stoichiometry to convert the given masses to moles. Finally, we can use the stoichiometric ratio to find the mass of HgS formed.
Using the molar masses of Hg(ClO3)2 (307.609 g/mol) and Na2S (78.045 g/mol), we can calculate the number of moles of each reactant:
- Hg(ClO3)2: 91.345 g / 307.609 g/mol = 0.297 mol
- Na2S: 13.180 g / 78.045 g/mol = 0.169 mol
From the balanced chemical equation, we can see that one mole of Hg(ClO3)2 reacts with one mole of Na2S to form one mole of HgS. Therefore, the number of moles of HgS formed is 0.169 mol (the lower of the two reactants).
Finally, we can calculate the mass of HgS formed using its molar mass of 232.64 g/mol:
Mass of HgS = 0.169 mol × 232.64 g/mol
= 39.37 g