[Steve88]

Im an idiot when it comes to chemistry so i need some help from everyone on this wonderful forum please.

Here is the scenario that i need help with please:

In the 1L outfall sample today, among other things, 134µg mercury(II) nitrate has been found. In order to extract the mercury which is in solution, you will add sodium sulphide. The ensuing reaction will allow the mercury to precipitate out as a solid. To ensure that all the mercury has been extracted, what mass (in grams) of precipitant must be collected from 16kL of outfall water?

I hope you can help me get to the answer.

[Schrödinger]

Ok, I'll give you hints that will help you get to the answer...

1. Write down the chemical formula of Mercury(II) nitrate.

2. How many moles of mercury can be precipitated from one mole of mercury(II) nitrate? (hint : use the chemical formula of solid mercury and that of mercury(II) nitrate)

3. 1L of outfall sample contains 134 µg mercury nitrate. So, how much of it is present in 16kL?

4. Final clue : convert mass of mercury nitrate to moles of mercury nitrate. Using hint 2, find moles of mercury. Convert moles of mercury to mass of mercury

Hope this helps 

[Steve88]

Thanks very much for the reply 

I think i might have the answers that im looking for but if you can please check them for me because like i said i am a complete idiot when it comes to chemistry.

1. HG(N03)2

2. (This is where i doubt myself) I researched and come up with this answer to question number 2: 324.5998 g/mol

3. 2144000

4. (This one i am clueless but i did some more research and came up with a figure i think is right) The figure i come up with is: 73.87249791536

I think these are right but i could also be horribly mistaken. If i can asked you to please check these and tell me if they are right, i would be eternally grateful