Question

Assume that 8.5 L of iodine gas are produced at STP according to the following balanced equation:

Answer 1

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

Answer 2

Answer:

1. 0.38
2. 0.38
3. 27

             

Explanation:

right on edge