If a solution containing 107.68 g of mercury(II) perchlorate is allowed to react completely with a solution containing 17.796 g of sodium sulfide, how many grams of solid precipitate will be formed?

This is a limiting reagent (LR) problem and you know that when amounts are given for BOTH reactants.

Hg(ClO4)2 + Na2S ==> HgS + 2NaClO4

1a. mols Hg(ClO4)2 = grams/molar mass = ?
b. mols Na2S = g/molar mass = ?

2a. Using the coefficients in the balanced equation, convert mols Hg(ClO4)2 to mols HgI.
b. Do the same and convert mols Na2S to mols HgS.

3. The mols HgS formed in 2a ia likely not to agree with mols HgS in 2b. The correct answer is ALWAYS the smaller number and the reagent responsible for that is the LR.

4. Using the smaller number convert mols HgS to g by
g = mols x molar mass. The solid ppt is HgS.

Well, we've got quite the chemistry party going on here! Mercury(II) perchlorate and sodium sulfide are having a wild dance-off, but we need to figure out how much solid precipitate will be formed.

To do that, we need to find the limiting reactant. The limiting reactant is like the party pooper who determines how much fun everyone can have. It's the reactant that gets used up first, putting a cap on the amount of product that can be formed.

To find the limiting reactant, we need to compare the number of moles of each reactant. We can calculate the number of moles using the molar masses of mercury(II) perchlorate and sodium sulfide.

The molar mass of mercury(II) perchlorate (Hg(ClO4)2) is approximately 336.59 g/mol, and the molar mass of sodium sulfide (Na2S) is approximately 78.04 g/mol.

By using the molar masses, we can convert the masses of each reactant to moles:

For mercury(II) perchlorate:

107.68 g / 336.59 g/mol = 0.3201 mol

And for sodium sulfide:

17.796 g / 78.04 g/mol = 0.228 mol

Now, we can take a closer look at the reactant ratio from the balanced chemical equation to see how much product is formed.

The balanced equation for the reaction is:

Hg(ClO4)2 + Na2S → HgS↓ + 2 NaClO4

From the equation, we can see that the ratio between Hg(ClO4)2 and HgS is 1:1. This means that for every 1 mole of Hg(ClO4)2, we get 1 mole of HgS.

So, if we have 0.3201 mol of Hg(ClO4)2, we will also have 0.3201 mol of HgS if it were the limiting reactant.

However, the ratio between Na2S and HgS is 1:1. So, if we have 0.228 mol of Na2S, we will only get 0.228 mol of HgS, which is less than the amount we could get from Hg(ClO4)2.

Therefore, sodium sulfide is the limiting reactant. This means that we will only get 0.228 mol of HgS precipitate.

To find the mass of the HgS precipitate, we can use the molar mass of HgS (232.66 g/mol):

0.228 mol × 232.66 g/mol = 52.63008 g

So, approximately 52.63 grams of solid HgS precipitate will be formed.

But hey, don't forget to clean up after the chemistry party! Safety first!

To find the mass of solid precipitate formed, we need to determine the limiting reactant first.

1. Convert the mass of mercury(II) perchlorate to moles:
- Molar mass of mercury(II) perchlorate (Hg(ClO4)2) = atomic mass of Hg (200.59 g/mol) + 4 * atomic mass of Cl (35.45 g/mol) + 8 * atomic mass of O (16.00 g/mol)
- Molar mass of Hg(ClO4)2 = 200.59 g/mol + 4 * 35.45 g/mol + 8 * 16.00 g/mol = 448.59 g/mol
- Moles of mercury(II) perchlorate = mass of Hg(ClO4)2 / molar mass of Hg(ClO4)2
- Moles of mercury(II) perchlorate = 107.68 g / 448.59 g/mol

2. Convert the mass of sodium sulfide to moles:
- Molar mass of sodium sulfide (Na2S) = 2 * atomic mass of Na (22.99 g/mol) + atomic mass of S (32.07 g/mol)
- Molar mass of Na2S = 2 * 22.99 g/mol + 32.07 g/mol = 46.00 g/mol
- Moles of sodium sulfide = mass of Na2S / molar mass of Na2S
- Moles of sodium sulfide = 17.796 g / 46.00 g/mol

3. Use the stoichiometric coefficients from the balanced equation to determine the limiting reactant:
The balanced equation for the reaction between mercury(II) perchlorate and sodium sulfide is:
Hg(ClO4)2 + Na2S -> HgS + 2NaClO4

From the balanced equation, we can see that the stoichiometric ratio between Hg(ClO4)2 to HgS is 1:1. This means that 1 mole of Hg(ClO4)2 reacts with 1 mole of HgS.
Similarly, the stoichiometric ratio between Na2S to HgS is 1:1. This means that 1 mole of Na2S reacts with 1 mole of HgS.

Compare the moles of Hg(ClO4)2 to Na2S to determine the limiting reactant. The reactant that produces fewer moles of HgS is the limiting reactant.

4. Calculate the moles of HgS formed based on the limiting reactant:
The limiting reactant is the one that produces fewer moles of HgS.
Let's assume that moles of Hg(ClO4)2 is the limiting reactant in this case.
Moles of HgS formed = Moles of Hg(ClO4)2 (from step 1)

5. Convert the moles of HgS formed to mass:
- Molar mass of HgS = atomic mass of Hg (200.59 g/mol) + atomic mass of S (32.07 g/mol)
- Molar mass of HgS = 200.59 g/mol + 32.07 g/mol = 232.66 g/mol
- Mass of solid precipitate = Moles of HgS formed * molar mass of HgS

Plug in the values:

Mass of solid precipitate = (107.68 g / 448.59 g/mol) * 232.66 g/mol
Mass of solid precipitate = 55.74 g

Therefore, approximately 55.74 grams of solid precipitate will be formed.

To determine the number of grams of solid precipitate formed in a reaction, we need to identify the balanced chemical equation between mercury(II) perchlorate and sodium sulfide. Let's call mercury(II) perchlorate "Hg(ClO4)2" and sodium sulfide "Na2S".

The balanced chemical equation for the reaction between Hg(ClO4)2 and Na2S is:

Hg(ClO4)2 + Na2S -> HgS + NaClO4

From the balanced equation, we can see that 1 mole of Hg(ClO4)2 reacts with 1 mole of Na2S to produce 1 mole of HgS.

First, let's calculate the number of moles of Hg(ClO4)2 and Na2S in the given masses:

Molar mass of Hg(ClO4)2 = 2(Molar mass of Hg) + 8(Molar mass of Cl) + 8(Molar mass of O)
= 2(200.59 g/mol) + 8(35.45 g/mol) + 8(16 g/mol)
= 401.18 g/mol

Number of moles of Hg(ClO4)2 = Mass / Molar mass
= 107.68 g / 401.18 g/mol
= 0.2687 mol

Molar mass of Na2S = 2(Molar mass of Na) + 1(Molar mass of S)
= 2(22.99 g/mol) + 1(32.07 g/mol)
= 87.06 g/mol

Number of moles of Na2S = Mass / Molar mass
= 17.796 g / 87.06 g/mol
= 0.2041 mol

From the balanced equation, the stoichiometric ratio between Hg(ClO4)2 and HgS is 1:1. Therefore, the number of moles of HgS formed will be equal to the number of moles of Hg(ClO4)2 reacted.

Number of moles of HgS = 0.2687 mol

To calculate the mass of HgS, we need to multiply the number of moles by its molar mass:

Molar mass of HgS = Molar mass of Hg + Molar mass of S
= 200.59 g/mol + 32.07 g/mol
= 232.66 g/mol

Mass of HgS = Number of moles of HgS * Molar mass of HgS
= 0.2687 mol * 232.66 g/mol
= 62.6 g

Therefore, approximately 62.6 grams of solid precipitate, namely HgS, will be formed in the reaction between the given quantities of Hg(ClO4)2 and Na2S.