If a solution containing 118.44 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?

How many grams of the reactant in excess will remain after the reaction?

Hg(ClO3)2 + Na2S ==> HgS + 2NaClO3

mols Hg(ClO3)2 = grams/molar mass = ?
mols Na2S = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols Hg(ClO3)2 to mols HgS.
Do the same to convert mols Na2S to mols HgS.
It is likely these two values will not be the same which means one of them must not be right; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Then g ppt = mols HgS (the smaller value) x molar mass HgS = ?

Then to find the amount of the other reagent in excess you must determine the amount of the other reagent used. To do that use the coefficients again, convert mols of the limiting reagent (LR) to mols of the other reagent and multiply that by the molar mass of the other reagent. That will give you the grams of the other reagent used. Subtract from the initial amount to find how much is left.

To answer this question, we first need to write out the balanced chemical equation for the reaction between mercury(II) perchlorate (Hg(ClO4)2) and sodium sulfide (Na2S):

Hg(ClO4)2 + Na2S ⟶ HgS + 2NaClO4

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

Step 1: Calculate the moles of Hg(ClO4)2 and Na2S used in the reaction.
To find the moles of a substance, you can use the formula:
moles = mass / molar mass

The molar mass of Hg(ClO4)2 is:
Hg = 200.59 g/mol
ClO4 = 35.45 + 16.00 * 4 = 99.45 g/mol
Molar mass of Hg(ClO4)2 = 200.59 + 2 * 99.45 = 399.49 g/mol

The moles of Hg(ClO4)2 used in the reaction:
moles of Hg(ClO4)2 = 118.44 g / 399.49 g/mol

The molar mass of Na2S is:
Na = 22.99 g/mol
S = 32.07 g/mol
Molar mass of Na2S = 2 * 22.99 + 32.07 = 87.04 g/mol

The moles of Na2S used in the reaction:
moles of Na2S = 17.796 g / 87.04 g/mol

Step 2: Determine which reactant is limiting and which is in excess.
To determine the limiting reactant, we compare the moles of each reactant from step 1.

From the balanced equation, we can see that the stoichiometry between Hg(ClO4)2 and Na2S is 1:1. So, whichever reactant has a smaller number of moles will be the limiting reactant, and the other reactant will be in excess.

Step 3: Calculate the moles of HgS formed.
Since the stoichiometry between Hg(ClO4)2 and HgS is also 1:1, the moles of HgS formed will be equal to the moles of the limiting reactant.

Step 4: Calculate the mass of HgS.
To find the mass of HgS formed, we use the formula:
mass = moles * molar mass

The molar mass of HgS is:
Hg = 200.59 g/mol
S = 32.07 g/mol
Molar mass of HgS = 200.59 + 32.07 = 232.66 g/mol

The mass of HgS formed:
mass of HgS = moles of HgS * 232.66 g/mol

Step 5: Calculate the moles and mass of the excess reactant remaining.
To find the moles and mass of the excess reactant remaining, we need to subtract the moles used in the reaction from the moles initially present.

moles of excess reactant = initial moles of reactant - moles used in the reaction
mass of excess reactant = moles of excess reactant * molar mass

Step 6: Calculate the mass of the excess reactant remaining in grams.
To convert the moles of excess reactant into grams, we use the formula:
mass = moles * molar mass

Now we can calculate the answers to the question.

Note: Make sure to round your final answers to the appropriate number of significant figures.

Answer:
The mass of solid precipitate (HgS) formed is [calculate the value using the given values and the steps described above].

The mass of the reactant in excess remaining after the reaction is [calculate the value using the given values and the steps described above].