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

3 answers

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.