Question

How much heat is liberated at constant pressure when 1.41 g of potassium metal reacts with 6.52 mL of liquid iodine monochloride (d = 3.24 g/mL)? 2K(s) + ICl(l) → KCl(s) + KI(s)

Answer 1

Answer:

The correct answer is -  13.33 kJ of heat

Explanation:

To know which one is the limiting reagent, determine the number of moles of each reagent in order .

n(K) = mass/atomic weight = 1.41/39 = 0.036 moles

Density of ICl = Mass/Volume

3.24 = Mass/6.52

Mass of ICl = 21.12 g

n(ICl) = mass/molar mass = 21.12/162.35 = 0.130 moles

2 moles of K reacts with 1 mole of ICl

0.036 moles of K will react with = 0.036/2 = 0.018 moles of ICl

since the amount of moles of ICl is more than 0.018, it is in excess and hence K is the limiting reagent. Now, use the balance equation to determine the amount of heat liberated:

2 moles of K gives out -740.71 kJ of heat

1 mole of K will give out = -740.71/2 = 370.36 kJ of heat

0.036 moles of K will give out = 0.036 × 370.36 = 13.33 kJ of heat

Thus, the correct answer is -  13.33 kJ of heat