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