Assume that 8.5 l of iodine gas are produced at stp according to the following balanced equation: 2kl+cl2-> 2kcl+i2 how many moles of i2 are produced? how many moles of cl2 are used? how many grams of cl2 are used?

Balanced chemical equation is as;

                                  2 Kl  +  Cl₂     →     2 KCl  +  I₂

1)  Moles of I₂ produced;

             22.4 L volume is occupied by   =  1 mole of I₂ at STP
So,
             8.5 L of I₂ will be occupied by  =  X mole of I₂ at STP

Solving for X,
                       X  =  (8.5 L × 1 mol) ÷ 22.4 L

                       X  =  0.379 Moles of I₂ 

2)  Moles of Cl₂ used;

                         22.4 L I₂ utilized  =  1 mole of Cl₂  gas
So,
                    8.5 L of I₂ will utilize =  X mole of Cl₂  gas

Solving for X,
                       X  =  (8.5 L × 1 mol) ÷ 22.4 L

                       X  =  0.379 Moles of Cl₂ 

3)  Grams of Cl₂ used;

As,
                       Moles  =  Mass / M.mass
Or,
                       Mass  =  Moles × M.mass

                       Mass  =  0.379 mol × 70.90 g.mol⁻¹

                       Mass  =  26.87 grams of Cl₂

Since there is the same amount of agn03 in both silver chloride and silver nitrate they produce the same amount of moles. therefore silver chloride produces the same amount of silver nitrate, so your answer is 15.0 moles of silver nitrate. i hope this !