Question

The alkali metals react with the halogens to form ionic metal halides. what mass of potassium chloride forms when 5.11 l of chlorine gas at 0.943 atm and 286 k reacts with 29.0 g potassium?

Answer 1

ideal gas law: PV = nRT so ..... V = PV/(RT) 

Initial number of moles of Cl, n = 0.943*5.11/(0.08206 × 286) mol = 0.2053 moles.

We know the molar mass of K (potassium) = 39.0 g/mol 
sooo....
The Initial number of moles of K = 29.0 g/(39.0 g/mol) = 0.7436 moles

Find the balanced equation for the reaction : 2K + Cl2 → 2KCl 
Mole ratio of K:Cl = 2:1 

So after the reaction, the amount of K needed = (0.2053 mol) × 2 = 0.4106 mol which is less than 0.7436 mol 

This means that K is in excess but Cl completely reacts. 

 So we know the mole ratio is  Cl:KCl = 1 : 2 

Number of moles of Cl (completely) reacted = 0.2053 mol which means the number of moles of KCl formed = (0.2053 mol) × 2 = 0.4106 mol 

Molar mass of KCl = (39.0 + 35.5) g/mol = 74.5 g/mol 
Mass of KCl formed = 0.4106 mol * 74.5 g/mol = 30.6 g