Finding the mass needed for a reactant if you know the mass of the product?

Ok, so you need to figure out the number of mols of each substance present and then basically solve for the x

2.56kg of I2 to mols, 2.54 kg of ICl3 to mols, you know that should be a 1:1 molar ratio since they have 2s on each sideof the equation. You need to figure out what number of mols of Cl2 will react with the mols of I2 to make ICl3, i'm assuming Cl2 should be your limiting reagent so you're figuring out how many mols of that it will take to make 2.54 kg of ICl3.

(126.9x2)=I2  Cl2=70.9, 233.25=ICl3 molar masses
g/mols=mm
g/mm=mols
I2=10.0866mols
ICl3=10.889mols
Cl2=?
This is approx. 1:1, Cl2 is going to be 3/2:1, I got about 16.334 mols for Cl2, multiply that by the molar mass (2x35.45) and that should give you the massof Cl2.

I reread your question and realized you gave the masses of Iodine trichloride twice, so the mass I used may haev been for iodine or chlorine, but that is how you would solve the problem given reactant and product.

Ok
first thing to do is fing the moles of ICl3
(sorry  i dont have a calc to hand so you can do this with instructions)
so add up the RFM of ICl3 so mass of iodine +(3 masses of chlorine)
Then divide the 2.54kg by the mass
so 2540/RFM
now you have number of moles.
Now you work out the ratio of iodine to ICl3 to do this you look at the amount of moles needed on the left to make 2 moles on the right
the answer is 2 as thats the number infront of the Iodine
so the ratio is 2:2 = 1:1.
this means the amount of moles needed of iodine = the amount of moles of product produced
so Moles ICl3 = Moles I2
so you then get the number of moles of ICl3 you worked out, and multiply it by the RFM of I2
this gives you in grams the amount of I2 needed