suppose that 2.00 mL of 0.01 mol/L sodium sulphide is used to test a 50.0 mL sample of water containing 0.0005 mol/L mercury nitrate ions. What mass of precipate is formed?
Na2 = Hg(NO3)2 --->2NaNO3 + HgS
My Work:
nNa2S=0.002*0.01
=0.0002 mol
nHg(NO3)2=0.05*0.0005
=0.000025mol
Na2S is limiting reactant
xHgS=0.00002mol
m=0.0002*56.371
=0.00113 g
therefore 1.13*10^-3 grams is formed
The answer you have is not correct but the method is ok. I have a few bold faced remarks in appropriate spots below.
suppose that 2.00 mL of 0.01 mol/L sodium sulphide is used to test a 50.0 mL sample of water containing 0.0005 mol/L mercury nitrate ions. What mass of precipitate is formed? A picky detail: I would have named it mercury(II) nitrate or mercuric nitrate so as to distinguish between the two usual mercury ions. Second, I would have omitted the word "ions" since you are testing for mercury(II) ions and not mercury(II) nitrate ions.
Na2 = Hg(NO3)2 --->2NaNO3 + HgS
Na2S + Hg(NO3)2 ==>
My Work:
nNa2S=0.002*0.01
=0.0002 mol
The correct number of mols is 2 x 10^-5. I think this is just a typo since you used the correct number below.
nHg(NO3)2=0.05*0.0005
=0.000025mol
Na2S is limiting reactant
xHgS=0.00002mol This is correct
m=0.0002*56.371 You omitted one of the zeros AGAIN AND you didn't use the correct molar mass for HgS.
=0.00113 g
therefore 1.13*10^-3 grams is formed
thanks very much!
You're welcome.
To solve this problem, let's first determine the limiting reactant by comparing the number of moles of Na2S and Hg(NO3)2.
The number of moles of Na2S can be calculated using the volume and concentration provided:
nNa2S = volume (in L) * concentration (in mol/L)
nNa2S = 0.002 L * 0.01 mol/L
nNa2S = 0.00002 mol
Similarly, the number of moles of Hg(NO3)2 can be calculated:
nHg(NO3)2 = volume (in L) * concentration (in mol/L)
nHg(NO3)2 = 0.05 L * 0.0005 mol/L
nHg(NO3)2 = 0.000025 mol
Comparing the number of moles, we can see that Na2S is the limiting reactant because it yields fewer moles (0.00002 mol) compared to Hg(NO3)2 (0.000025 mol).
Now, we can determine the number of moles of HgS formed by using the balanced equation:
Na2S + Hg(NO3)2 → 2NaNO3 + HgS
From the balanced equation, we can see that the ratio between Na2S and HgS is 1:1. Since we have 0.00002 mol of Na2S, we will also have 0.00002 mol of HgS.
Next, we can calculate the mass of HgS using its molar mass:
m = number of moles * molar mass
m = 0.00002 mol * (2 * atomic mass of Hg + atomic mass of S)
m = 0.00002 mol * (2 * 200.59 g/mol + 32.06 g/mol)
m ≈ 0.00113 g
Therefore, approximately 0.00113 grams of precipitate (HgS) is formed in this reaction.