[Evaldas]

Hi there! My name is Evaldas, and I'm 16, for this summer I've decided to get better at chemistry, because I've decided to study medicine after school. I bought a few books, and the most questions I have are about Molar mass (and everything that goes with it), and oxidation-reduction. I hope you can help me . So the first question (it's an exercise from one book, I wasn't able to find the same exact example, so I thought "why not ask here?".
So:
How many grams and moles will be produced (I'm translating it literally!) of plumbum sulphide, if 69 grams of plumbum is reacting with sulphur?
The formula of plumbum sulphide is Pb^2+S^2- (if you must know )
I'd really appreciate if you'd explain me how to solve this using formulas like n=m/M, m=n x M, N = n x NA (Avogrado's constant - 6.02 x 10^23)
By the way in Lithuania when we count atomic mass of a compound we don't use the whole long atomic weights of every  element. We round up, so for example if in the table it says that H weights 1.00794, we use 1, and the only exception is Chlorine, it's 35.5.

After you help me, nice people, with this question, there's another coming up . I hope I'm not too much trouble.

[sjb]

I'm sure that you really do use "long atomic weights of every  element" if necessary, but this may only confuse at an early stage of education.

Can you write a balanced equation for formation of PbS from lead (plumbum) and sulphur?

[Evaldas]

Balanced? Does it need balancing? I think it's just:

Pb           +   S > PbS
0.33 mol

M(Pb)=207 g/mol.
M(S)=32 g/mol
so as m(Pb)=69g then n(Pb)=207/69=0.33 mol.

By the way the answer is m=79.7 g and n=0.3 mol, just don't know how to get to that answer

[sjb]

So you have 0.33 mol of lead, how much PbS can you make, if you have an excess of sulfur?

[Evaldas]

I can make 0.33 mol of PbS, no? 

[sjb]

Right, and the mass of that is..?

Not 100% sure why it says 0.3 mol, perhaps just a rounding thing?

[Evaldas]

But why's that right? I just guessed...

Wait, let me get my beloved periodic table 
And the m_(PbS)=n(PbS) x M(PbS) = 0.33 mol. x 248 g/mol. = 81.84 g.
because M_(PbS)=207+32=248 g/mol.


Yeah, I guess just a rounding thing  , I think the most exact would be 23/69 mol.

[Evaldas]

Oh wait, I remember why that is correct:
Because in PbS we have one atom of S, so:
n_(PbS)=n_(S)
If we had for example 3S then it would be:
n_(S)=3n_(PbS3) (I know that the the formula is not right, it's just an example!)

[Evaldas]

So the answer is And the m_(PbS)=n_(PbS) x M_(PbS) = 0.33 mol. x 248 g/mol. = 81.84 g.

Not 79.7g? (as is says in the book)

[Borek]

207+32=248

You sure?

[Evaldas]

Oh my!  Oh no...  Sorry!
207+32=239
then m_(PbS) = n_(PbS) x M_(PbS) = 0.33 mol. x 239 g/mol = 78.87 g.
But still not 79.7 g...

[UG]

That is just a rounding error. If you did it like, n(Pb) = 69 / 207.19 = 0.3330276558 moles
m(PbS) = 0.3330276558 x (207.19 + 32.064) = 79.678 g ~ 79.7 g 

[Evaldas]

Oh, ok that's great!
One more question about that same exercise:
How do I write it all down?
I know that the reaction is Pb + S  PbS
But where do I put the mass of Pb (69g)? Above Pb or below..? Or should I write moles in the reaction not mass?

[UG]

Well I don't think you have to write the numbers anywhere in the equation, I have never done it.

[Evaldas]

Ok...  Thanks.

So that was exercise number 1.
And here's exercise number 2 (I can't understand this one too)
Count the mass in grams of a) 1 atom of oxygen; b) 2 nitrogen atoms; c) 3 sulphur atoms; d) 1 water molecule; e) 2 NH₃ molecules

Now again, I know that the answers are:
a) 2.7 x 10^-23 g; b) 4.6 x 10^-23 g; c) 16 x 10^-23 g; d) 3 x 10^-23 g; e) 5.6 x 10^-23 g

But how do I get these numbers?

And is this right: 2.7 x 10^-23 =1 / 2.7 x 10²³ ?