Question

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

Answer 1

<b>The answer you have is not correct but the method is ok. I have a few bold faced remarks in appropriate spots below.</b>

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? <b>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.</b>
Na2 = Hg(NO3)2 --->2NaNO3 + HgS

<b>Na2S + Hg(NO3)2 ==> </b>

My Work:
nNa2S=0.002*0.01
=0.0002 mol
<b>The correct number of mols is 2 x 10^-5. I think this is just a typo since you used the correct number below. </b>

nHg(NO3)2=0.05*0.0005
=0.000025mol

Na2S is limiting reactant

xHgS=0.00002mol <b>This is correct</b>

m=0.0002*56.371 <b>You omitted one of the zeros AGAIN AND you didn't use the correct molar mass for HgS.</b>

=0.00113 g

therefore 1.13*10^-3 grams is formed

Answer 2

thanks very much!

Answer 3

You're welcome.