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
the limiting reactant is the mercury nitrate.
c. calculate the maximum mass of precipitate formed.
what is the precipitate?
Very confused with this question
Mercury sulfide
Chemical Compound
Mercury sulfide, mercuric sulfide, mercury sulphide, or mercury sulfide is a chemical compound composed of the chemical elements mercury and sulfur. It is represented by the chemical formula HgS.
"******It is virtually insoluble in water.*****"
Wikipedia
Formula: HgS
Molar mass: 232.66 g/mol
Density: 8.10 g/cm³
Melting point: 1,076°F (580°C)
I can understand why this question might be confusing, but I'll guide you through it step by step.
a. To start, we need to write the balanced equation for the reaction between sodium sulphide (Na2S) and mercury (II) nitrate (Hg(NO3)2). The molecular formula of the products can be determined by swapping the anions of the reactants. Therefore, the balanced equation is:
Hg(NO3)2 + Na2S -> HgS + 2NaNO3
b. To determine the limiting reactant, we compare the moles of each reactant in the given volume of the solution.
For Na2S:
Volume = 2.00 mL = 2.00 mL / 1000 mL = 0.00200 L
Concentration = 0.02 mol/L
Moles = Concentration * Volume = 0.02 mol/L * 0.00200 L = 0.00004 mol
For Hg(NO3)2:
Volume = 60.0 mL = 60.0 mL / 1000 mL = 0.0600 L
Concentration = 0.0004 mol/L
Moles = Concentration * Volume = 0.0004 mol/L * 0.0600 L = 0.000024 mol
As you calculated, the moles of Na2S (0.00004 mol) are greater than the moles of Hg(NO3)2 (0.000024 mol). Therefore, the limiting reactant is Hg(NO3)2.
c. To calculate the maximum mass of precipitate formed, we need to use stoichiometry. The balanced equation tells us that the stoichiometric ratio between Hg(NO3)2 and HgS is 1:1.
From the limited reactant, we can determine the moles of HgS formed, which will be the same as the moles of Hg(NO3)2 used:
Moles of HgS = 0.000024 mol
To convert moles of HgS to grams, we need to use the molar mass of HgS. The molar mass of HgS is the sum of the atomic masses of mercury (Hg) and sulfur (S), which can be found on the periodic table.
Molar mass of HgS = (molar mass of Hg) + (molar mass of S)
Using the molar masses:
Molar mass of HgS = (200.59 g/mol) + (32.07 g/mol) = 232.66 g/mol
Now, we can calculate the mass of HgS formed:
Mass of HgS = Moles of HgS * Molar mass of HgS
= 0.000024 mol * 232.66 g/mol
This will give you the maximum mass of precipitate formed.
I hope this clears up the confusion. Let me know if you have any further questions!