[Borek]

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

It depends on how these things are taught, I suppose in Lithuania they may use similar system to the one used in Poland:

http://www.chembuddy.com/?left=balancing-stoichiometry&right=ratio-proportions

[Evaldas]

About how we write it in Lithuania:
For example if we get an exercise, like this one "How many moles and grams of oxygen there is in 1 mol of H₂SO₄?"
We write 'a shortened' version of the exercise, we write what's given, what to find and solution and the answer.
So for this one we would write:
Given:
n_(H2SO4) = 1 mol.
What to find:
m_(O) - ?
n_(O) - ?
Solution:
M_(H2SO4) = 1 x 2 + 32 + 4 x 16 = 98 g/mol.
m_(H2SO4) = M_(H2SO4) x n_(H2SO4) = 98 g
M_(O) = 16 g/mol.
n_(O) = 4n_(H2SO4) = 4 x 1 mol. = 4 mol.
m_(O) = M_(O) x n_(O) = 16 g/mol x 4 mol = 64 g.
Answer:
There are 98 g, and 4 mol. of oxygen in H₂SO₄

So I guess it's something similar with Pb + S  PbS, I think you have to put in the mass or the moles of lead somewhere.

[UG]

1 atom of oxygen

One mole of oxygen weighs 16 grams = 6.02 x 10²³ oxygen atoms weighs 16 grams

So one atoms weighs 16/(6.023 x 10²³)

[Evaldas]

Great! I get it now

[Evaldas]

Ok, another one... I hope I'm not too much of a headache for you! As I said I just want to get better at chemistry during the summer...
I'm gonna try to solve it as I understand it, and I'm gonna later say what's unclear...
Exercise number 3.
What amount and mass of oxygen would evaporate when heating 442 g of Au₂O₃?
Given: 2Au₂O₃ 4Au + 3O₂
m_(Au2O3)=442 g
What to find:
n_(O) - ?
m_(O) - ?
Solution:
M_(Au2O3) = 2 x 197 + 3 x 16 = 442 g/mol.
n_(Au2O3) = 1 mol.
n_(O)=3n_(Au2O3)= 3 x 1 mol. = 3 mol.
m_(O)=M_(O) x n_(O) = 16 g/mol x 3 mol = 48 g
Answer: 3 mol. and 48 g of oxygen will evaporate when heating 442 g of Au₂O₃
(By the way the answer does not exactly match to what it says in the book: there it says 48 g. and 1.5 mol., where did I mess up? Maybe it should be - n_(O)=1.5n_(Au2O3)= 1.5 x 1 mol. = 1.5 mol., because there's 2 oxygens at one side and 3 oxygens at other?))

But why this one's unclear? I don't understand when I heat Au₂O₃ all the oxygens evaporate? None stay in the compound? So what if I heat gold(III) oxide, I just get pure gold?
By the way in the exercise it says 'heated', should there be a *t above the arrow? Would that change something?

[UG]

There is 1 mole of Au₂O₃ in 442 grams. The mole ratio Au₂O₃: O₂ is 2 : 3 therefore the amount of oxygen is 1 x 3/2 = 1.5 moles of diatomic oxygen.

'So what if I heat gold(III) oxide, I just get pure gold?'
Yes, I believe so.

[Evaldas]

But then m_(O) = 16 g/mol x 1.5 mol = 24 g..?  There, in the book it says 48 g 

[UG]

But then m_(O) = 16 g/mol x 1.5 mol = 24 g..?  There, in the book it says 48 g 

That should be m_(O2) = 32 x 1.5

[Evaldas]

Ah, ok, great! Thanks! I just need a few moments to diggest that information in my head...

[Evaldas]

Ok and another one. This one's pretty interesting. It's from a test we did in class. Now I don't have the answer for this one, so I must ask about it. It's and exercise with 6 questions, I just don't know the last one.
Exercise 4:
When the acidity in our stomaches is higher than normal, we feel pain, which we can soothe with medicine (for example "Maalox" or "Rennie" pill)
In one "Maalox" pill there is 400 mg of aluminum hydroxide Al(OH)₃, 400 mg of magnesium hydroxide Mg(OH)₂ and flavourants.
1) Q: What acid is found in human's stomach? A: HCl
2) To which group the named compounds in the pills belong?
Underline the correct answer:
a) acids b) bases
3) Q: "Maalox" neutralizes the acidity in the stomach. What does it mean neutralizes?
A: OH^- ions join H⁺ ions and form H₂O.
4) Q: How the pH of gastric juices will change after taking these pills?
A: It will get higher.
5) Q: Write down the reaction equations, that happen in the stomach after taking "Maalox" pill.
A: 1) Mg(OH)₂ + 2HCl_(aq)    MgCl₂ + 2H₂O
2) Al(OH)₃ + 3HCl_(aq)  AlCl₃ + 3H₂O
6) And here's the one I can't solve:
Count how many grams of HCl in gastric juices would react after taking two "Maalox" pills.


I really have no idea what to do with it. Please, direct me to the right direction...

[sjb]

6) And here's the one I can't solve:
Count how many grams of HCl in gastric juices would react after taking two "Maalox" pills.


I really have no idea what to do with it. Please, direct me to the right direction...

400 mg of Al(OH)₃ is how many moles? Which react with how many moles of HCl, according to your balanced equation. What is the mass of that amount of HCl? Repeat for Mg(OH)₂. This is for 1 pill, so for 2 pills?

[Evaldas]

Let's see...
n_(Al(OH)3) = 0.4 g / 78 g/mol = 0.005 mol.
n_(HCl) = n_(H) = n_(Cl) = 3n_(Al(OH)3) = 3 x 0.005 mol = 0.015 mol? 
m_(HCl) = 36.5 g/mol x 0.015 mol = 0.54 g

now
n_(Mg(OH)2) = 0.4 g / 58 g/mol = 0.006 mol.
n_(HCl) = n_(H) = n_(Cl) = 2n_(Mg(OH)2) = 2 x 0.006 mol = 0.012 mol.
m_(HCl) = 36.5 g/mol x 0.012 mol = 0.43 g

So as there are two pills I need to multiply everything by two?
Then:
m_1(HCl) = 0.54 g x 2 = 1.08 g
m_2(HCl) = 0.43 g x 2 = 0.86 g
m_(HCl) = 1.04 + 0.86 = 1.94 g

I bet, I messed up both times counting the n_(HCl), but I don't know where! Or didn't I?

[sjb]

I think you have "messed up" slightly in carrying forward figures that have been rounded prematurely, but the essence of the maths is correct.

[Evaldas]

But the thing is that I don't get WHY that is correct  , I just did those things 'mechanically' and that's it...

[Evaldas]

There is 1 mole of Au₂O₃ in 442 grams. The mole ratio Au₂O₃: O₂ is 2 : 3 therefore the amount of oxygen is 1 x 3/2 = 1.5 moles of diatomic oxygen.

'So what if I heat gold(III) oxide, I just get pure gold?'
Yes, I believe so.

So back to exercise 3:
Why is the ratio like that? I think it should be the same on both sides, cos the exact same amount of oxygen atoms on one side join into molecules of 2 atoms, and on the other into molecules of 3...