One method of preparing clo2 is by the reaction of chlorine and sodium chlorite: cl2 (g) + 2 naclo2 (s) → 2 clo2 (g) + 2 nacl (s) if you allow 15.0 g of naclo2 to react with 2.00 l of chlorine gas (cl2) at a pressure of 1.50 atm and at 21 oc, how many grams of clo2 can be prepared?

            5.59 g of ClO₂

Solution:

First calculate moles of Cl₂:

According to ideal gas equation,

                                                  P V  =  n R T
Or,
                                                  n  =  P V / R T
Putting values,
                         n  =  (1.5 atm × 2.0 L) ÷ (0.0821 atm.L.mol⁻¹.K⁻¹ × 294 K)

                         n  =  0.124 mol of Cl₂

According to equation,

                         1 mole Cl₂ required  =  2 moles of NaClO₂
So,
              0.124 mole Cl₂ will require  =  X moles of NaClO₂

Solving for X,
                       X  =  (0.124 mol × 2 mol) ÷ 1 mol

                       X  =  0.248 mol of NaClO₂

And given mass of NaClO₂ equals followinf moles,

                       n  =  15 g ÷ 90.45 g.mol⁻¹

                       n  =  0.1658 mol of NaClO₂

It means we are given with less amount of NaClO₂, hence it is limiting reagent and will control the amount of product formed.
So, According to equation

                    2 moles of NaClO₂ produced  =  67.45 g of ClO₂
So,
       0.1658 moles of NaClO₂ will produce  =  X g of ClO₂

Solving for X,
                       X  =  (67.45 g × 0.1658 mol) ÷ 2 mol

                       X  =  5.59 g of ClO₂

The correct answer is "atomic weight".

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