# Suppose that 2.00mL of 0.02 mol/L aqueous sodium sulphide, Na2S is used to test a 60.0mL sample of water containing 0.0004 mol/L mecury (II) nitrate Hg(NO3)2, ions.

a. Write the balanced equation

Hg(NO3)2 + Na2S = HgS + 2NaNO3

b. Determine the limiting reactant

Na2S = 2.00mL/1000mL x 0.02mol/L

= 0.00004 mol

Hg(NO3)2 = 60.0mL/1000mL x 0.0004mol

= 0.000024 mol

limiting reactant is the mercury nitrate.

c. calculate the maximum mass of precipitate formed.

what is the precipitate?

## You need to learn the solubility rules. HgS is the ppt (the solid).

## The precipitate formed in this reaction is mercury(II) sulfide (HgS).

To calculate the maximum mass of precipitate formed, we need to use stoichiometry.

From the balanced equation, we can see that the ratio of Hg(NO3)2 to HgS is 1:1. This means that every 1 mole of Hg(NO3)2 reacts to form 1 mole of HgS.

Given that the limiting reactant is Hg(NO3)2 with 0.000024 mol, we can calculate the moles of HgS formed:

Moles of HgS = 0.000024 mol

Now, we need to convert the moles of HgS to grams using the molar mass of HgS. The molar mass of HgS is the sum of the molar masses of mercury (Hg) and sulfur (S):

Molar mass of HgS = atomic mass of Hg + atomic mass of S

= (200.59 g/mol) + (32.06 g/mol)

= 232.65 g/mol

Now, we can calculate the mass of HgS using the formula:

Mass of HgS = Moles of HgS x Molar mass of HgS

= 0.000024 mol x 232.65 g/mol

= 0.00558 g

Therefore, the maximum mass of precipitate formed (HgS) is 0.00558 grams.

## The precipitate formed in this reaction is HgS (mercury(II) sulfide).

To calculate the maximum mass of precipitate formed (HgS), we need to know the molar mass of mercury(II) sulfide. The molar mass of HgS is:

Mercury (Hg) = 200.59 g/mol

Sulfur (S) = 32.06 g/mol

So, the molar mass of HgS = 200.59 g/mol + 32.06 g/mol = 232.65 g/mol

Now we can calculate the maximum mass of precipitate (HgS) formed using the limiting reactant (Hg(NO3)2):

Given moles of Hg(NO3)2 = 0.000024 mol

Molar mass of HgS = 232.65 g/mol

Mass of HgS = moles of Hg(NO3)2 x molar mass of HgS

= 0.000024 mol x 232.65 g/mol

= 0.00556 g

Therefore, the maximum mass of precipitate (HgS) formed is 0.00556 grams.