Cl2+3F2 -> 2ClF3

how many grams of ClF3 form when .204 moles if F2 react with excess Cl2?

Using equation Cl2+3F2----->2ClF3

How many grams of ClF3 form when 0.204 moles of F2 react with excess Cl2?

To determine the grams of ClF3 formed when 0.204 moles of F2 react with excess Cl2, you need to use the given balanced chemical equation:

Cl2 + 3F2 -> 2ClF3

From the balanced equation, you can see that 1 mole of F2 reacts to form 2 moles of ClF3. So, you need to convert the moles of F2 to moles of ClF3 using the stoichiometry.

0.204 moles of F2 x (2 moles of ClF3 / 3 moles of F2) = 0.136 moles of ClF3

Now, to calculate the grams of ClF3 formed, you'll need the molar mass of ClF3. From the periodic table, the molar masses are:

Molar mass of Cl = 35.45 g/mol
Molar mass of F = 18.998 g/mol

So, the molar mass of ClF3 = (35.45 g/mol) + (3 x 18.998 g/mol) = 97.454 g/mol

Finally, you can find the grams of ClF3 formed by multiplying the moles of ClF3 by its molar mass:

0.136 moles of ClF3 x 97.454 g/mol = 13.26 grams of ClF3

Therefore, when 0.204 moles of F2 react with excess Cl2, 13.26 grams of ClF3 will be formed.