2.00mL of 0.01 mol/L aqueous sodium sulphide is used to test a 50.omL sample of water containing 0.0005 mol/L mercury (II) nitrate ions. What mass of precipitate is formed?
Na2S + Hg(NO3)2 --- 2NaNo3 + HgS
determine limiting reactant
using limiting reactant, calculate the masimum mass of precipitate formed
mol Na2S = (0.01mol/L)*2.00ml*1L/1000mL = 0.002mol
mol Hg(NO3)2 = 50.0ml*(0.0005mol/L)*1L/1000mL = 0.000025=2.5*10^-5
Hg(NO3)2 limiting reagent
max mass of precipitate = 2.5*10^-5 * molar mass of Hg(NO3)2
I thought I was to use a formula n=cXv. I had NaS
c=.002 L
V >01 mol/L
Therefore I got NaS as .00002. So I had Na S as the limiting reagent
moles = M x L
For Na2S, 0.01 x 0.002 = 2 x 10^-5 moles
moles Hg(NO3)2 =
0.0005 x (50/1000)= 2.5 x 10^-5 moles
Limiting reagent is Na2S. (The set up for Na2S by bun is correct but math is not; answer is 2 x 10^-5 moles and not 2 x 10^-3 moles.
To determine the limiting reactant, we need to compare the number of moles of each reactant and see which one is present in the least amount.
First, let's calculate the number of moles of sodium sulphide (Na2S) in the 2.00 mL of the 0.01 mol/L solution:
Moles of Na2S = Volume (L) x Concentration (mol/L)
Moles of Na2S = (2.00 mL / 1000 mL/L) x 0.01 mol/L
Moles of Na2S = 0.00002 mol
Next, let's calculate the number of moles of mercury (II) nitrate (Hg(NO3)2) in the 50.0 mL water sample:
Moles of Hg(NO3)2 = Volume (L) x Concentration (mol/L)
Moles of Hg(NO3)2 = (50.0 mL / 1000 mL/L) x 0.0005 mol/L
Moles of Hg(NO3)2 = 0.000025 mol
Since the number of moles of Hg(NO3)2 is smaller than Na2S, Hg(NO3)2 is the limiting reactant.
Now, let's calculate the maximum mass of precipitate (HgS) formed using the limiting reactant (Hg(NO3)2). We can use the balanced equation to calculate the ratio of moles of HgS to moles of Hg(NO3)2, and then multiply it by the molar mass of HgS to get the mass.
From the balanced equation, we see that the ratio of moles of HgS to moles of Hg(NO3)2 is 1:1.
Molar mass of HgS = atomic mass of Hg (200.59 g/mol) + atomic mass of S (32.07 g/mol)
Molar mass of HgS = 232.66 g/mol
Therefore, the maximum mass of precipitate formed is:
Mass of HgS = Moles of Hg(NO3)2 x Molar mass of HgS
Mass of HgS = 0.000025 mol x 232.66 g/mol
Mass of HgS = 0.0058165 g
So, the mass of precipitate formed is approximately 0.0058 g (rounded to four decimal places).