[Kate]

Hi.
Well, I need to know what's the mass equivalent to 45 cm3 of oxygenated water at 20 volume.

Here's what I know:

Molar volume = V/n

In NTP conditions, molar volume=22,4 dm3/mol

2H2O2 -> 2H2O + O2

1 mol H2O2 ------ 22,4 dm3
2 mol H2O2 ------ 44,8 dm3

I have no clue as to what to do next. Can someone help me with this ? Thanks.

[AWK]

45 cm3 of oxygenated water at 20 volume.

What does it mean?

[Kate]

Hydrogen peroxide at 20 volume means that when 1 volume of hydrogen peroxide is decomposed, it produces 20 volumes of O2.

[AWK]

If so, you can get 900 cm³ of O₂ at certain conditiond (eg STP)
the you have 900/22400 moles or 900 x 32 /22400 grams

[Kate]

Ok. So,

2 mol H2O2 ---- 22,4 dm3
             x ----- 20 dm3

x= 1,7857 mol

1,7857 mol --- 1 dm3
       x mol --- 45x10-3 dm3

x= 0,08 mol

M(H2O2)=34g/mol
m=2,72 g

Right ?

[AWK]

From your data I assume you decompose 45 cm³ of H₂O₂.
in this case your calculations are wrong

[Kate]

That's the way my teacher solved the problem. I got to admit, I don't really understand her reasoning on this particular problem either.

If this is wrong, (in your opinion) what's the result ?

[Borek]

x ----- 20 dm3

Why 20 dm³?

[Kate]

I think it's because of the oxygenated water at 20 volume.

[Borek]

As you have already told 20 volume means that 1 volume of water produces 20 volumes of gas when decomposing. Now, you have started with 45 mL of water - what volume of gas should be produced?

[Kate]

It should be produced 900 cm3 of gas.

[Borek]

Right - so now repeat your calculations for 900 mL, not 20 dm³.

[Kate]

So,

2 mol H2O2 ---- 22,4 dm3
             x ----- 0.9 dm3

x= 0,08 mol

M(H2O2)=34g/mol
m=2,72 g

So my teacher's calculations are wrong, correct ? Though the result/outcome is exactly the same.