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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?

Respuesta :

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
pstnonsonjoku

The mass of  potassium chloride formed is 31.5 g

The number of moles of chlorine gas  is obtained from the ideal gas equation;

P = 0.943 atm

V = 5.11 l

T = 286 K

n = ?

R = 0.082 atm L K-1mol-1

From;

PV = nRT

n = PV/RT

n =  0.943 atm * 5.11 l/0.082 atm L K-1mol-1 * 286 K

n =4.819 /23.452

n = 0.21 moles of Cl2

Number of moles of K = mass/molar mass = 29.0 g/39 g/mol = 0.74 moles

Equation of the reaction is;

2K + Cl2 ----> 2KCl

Since the reaction is 2:1

1 mole of Cl2 reacts with 2 moles of K

0.21 moles of Cl2 reacts with 0.21 * 2/1 = 0.42 moles of K

This means that K is the reactant in excess.

1 mole of Cl2 yields 2 moles of KCl

0.21 moles of Cl2 yields  0.21 * 2/1 = 0.42 moles of KCl

Molar mass of KCl = 75 g/mol

Mass of KCl = 0.42 moles of KCl * 75 g/mol

Mass of KCl = 31.5 g of KCl

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