in this Equation:Mg+F2--->MgF2 the mol Ratio is
1:1:1?

thanks.

Yes, that's correct.

Thank you :)

This is an interesting reaction to see in fact.

You light the magnesium (burns with white dazzling flames), but it in a jar of fluorine (be careful fluorine is irritating and toxic! ) and you'll see it burn in the gas, forming white crystals on the inner surface of the gas jar.

:p

Well, of course, this should be carried out in a fume cupboard and with mask if possible. But there are so many 'magical' reactions :)

Lol well I not going to try it alone :) thank you for give me this Information.

But the answer is like this if is known that 12 gr of Mg are Came into contact with 1.9 gr F2.

So I asked to find how many gr of MgF2 are Created in the end of the Process.

So I know that the mol Ratio is
1:1:1 so I did table and I know that 0.05 mol of F2 are Responded and also 0.05 mol of Mg are Responded. The answer can be that in the Process Created 39.215 gr of MgF2?
Something here not look me correct.

Thank you again.

12 g of Mg is 1 mol of Mg
1.9 g of F2 is 0.05 mol of F2

right.

Mass of MgF2 = 24 + 19 + 19 = 62

Mass of 0.05 mol of MgF2 = 0.05 x 62 = 3.1 g

[Note, maybe you're using more precise values for the relative masses of Mg and F. I don't have a periodic table right now nearby to get more accurate]

OH yes yes I did the same way but I Wrong in my Calculate now I did again and is gave me the same result like yours
Thanks

What kind of chemistry expert doesn't have a period table nearby at all times! ;)

Lol but still he was close is 62.2 or 62.3

When it's near to midnight and everybody else is asleep, or when there is an interesting show being broadcast in a few minutes ;)

Hi. I got more tow Section.
1) I asked to find how many mol of F2 are Participated In the reaction. My answer is 0.05 mol.

2)I asked to find Which material ( form the Responding side) are Remains in surplus and in what gr it was.

SO the material is Mg and the gr is 10.9.

I not show all my Calculate here Because it will make a real Mess.
So do I correct?
Thanks

Good, now the last part.

I'm getting 10.8 (or 10.785 yes, I have my periodic table now ;) )

Ar of Mg = 24.3

0.05 mol Mg = 1.215 g

Initial mass of Mg = 12 g

Mass in surplus = 12 g - 1.215 g = 10.785 g = 10.8 g

:)

yes I think you use in more NO after the point. :)
and what you mean Ar of Mg=24.3 what is Ar? Maybe you mean mM?
thank you.

by the way I did it not like you.

my way is like this: I find that there in this Process 0.5 mol of Mg that Entered to the system,
and in the end of the Process I know there is 0.05 mol of Mg that react.
SO I did 0.5-0.05=0.45
Then I did 0.45*24=10.8 gr

Ar is in fact written as A_r (the r is in subscript, or smaller and below the line). This stands for:

Relative atomic mass

Another thing I might use, is Mr, M_r

which stands for: Relative Molecular mass, or Relative formula mass, depending on the substance.

by the way i did it not like you.

my way is like this: i find that there in this Process 0.5 mol of Mg that Entered to the system,
and in the end of the Process i know there is 0.05 mol of Mg that react.
SO i did 0.5-0.05=0.45
Then i did 0.45*24=10.8 gr

Oh, this works as you can see. So far you understand what you are doing, it's good :)