A water-soluble salt formed by Hg2+ and chlorate 105.27 g was dissolved in a solution containing 16.642 g of sodium sulfide, which gave rise to an insoluble mercury salt.
How much insoluble mercury salt is formed?
How much of the excess reagent remains?
Hg(ClO3)2 + Na2S ==> HgS + 2NaClO3
mols Hg(ClO3)2 = g/molar mass = 105.27/367.49 = 0.28646
mols Na2S = 16.642/78.045 = 0.21324
Since the reactants are 1:1 then 0.21324 is the smaller moles so Na2S will be the limiting reagent (LR) and Hg(ClO3)2 will be the excess reagent (ER).
mols HgS formed = 0.21324 x (1 mol HgS/1 mol Na2S) = 0.21324 mols HgS formed. Grams HgS = mols HgS x molar mass HgS = ?
How much Hg(ClO3)2 is used by the 0.21324 mols Na2S. That's
mols Hg(ClO3)2 used = mols Na2S x (1 mol HgClO3)2/1 mol Na2S) = 0.21324 mols Na2S x (1/1) = 0.21324
0.28646 = mols Hg(ClO3)2 to start
-0.21324 = mols used
----------------------------
0.07322 mols Hg(ClO3)2 remaining.
grams remaining = mols remaining x molar mass = ?
Post your work if you still have questions.
To determine how much insoluble mercury salt is formed and how much of the excess reagent remains, we need to use the concept of limiting reagent.
1. Calculate the molar mass of the given compounds:
- The molar mass of chlorate (ClO3-) is:
Cl = 35.45 g/mol
O = 16.00 g/mol
Molecular mass of ClO3- = 35.45 + (3 * 16.00) = 97.45 g/mol
2. Convert the given mass of chlorate to moles:
Number of moles of chlorate = mass / molar mass = 105.27 g / 97.45 g/mol
3. Calculate the molar mass of sodium sulfide (Na2S):
Na = 22.99 g/mol
S = 32.07 g/mol
Molecular mass of Na2S = (2 * 22.99) + 32.07 = 87.05 g/mol
4. Convert the given mass of sodium sulfide to moles:
Number of moles of sodium sulfide = mass / molar mass = 16.642 g / 87.05 g/mol
5. Use the balanced chemical equation between Hg2+ and chlorate to determine the stoichiometry:
Hg2+ + 2ClO3- → Hg(ClO3)2
From the equation, we can see that 1 mole of Hg2+ reacts with 2 moles of ClO3- to form 1 mole of Hg(ClO3)2.
6. Determine the limiting reagent:
Compare the moles of chlorate (from step 2) and sodium sulfide (from step 4). The reactant with the smaller number of moles is the limiting reagent. Whichever reagent is in excess, any extra moles will not react.
7. Calculate the moles of insoluble mercury salt formed:
Since 1 mole of Hg2+ reacts with 2 moles of ClO3-, the moles of insoluble mercury salt formed will be half the number of moles of chlorate.
8. Calculate the mass of insoluble mercury salt formed:
Mass of insoluble mercury salt = moles of insoluble mercury salt formed * molar mass of Hg(ClO3)2
9. Calculate the moles of excess reagent remaining:
To calculate the moles of excess reagent remaining, find the difference between the initial moles of the excess reagent and the moles used in the reaction.
Initial moles of excess reagent = moles of sodium sulfide (from step 4) - (2 * moles of insoluble mercury salt formed)
10. Calculate the mass of excess reagent remaining:
Mass of excess reagent remaining = moles of excess reagent remaining * molar mass of sodium sulfide
By following these steps and plugging in the given values, you should be able to calculate the amount of insoluble mercury salt formed and the amount of excess reagent remaining.