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)

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

MrRoyal MrRoyal · Answer 2 · 2022-01-02T17:24:31+01:00

The amount of heat liberated at constant pressure is -13.33 kJ

The given parameters are:

Mass of potassium metal = 1.41g
Amount of liquid iodine monochloride = 6.52 mL

Start by calculating the number (n) of moles of each reagent using:

[tex]n = \frac{Mass}{Atomic\ weight }[/tex]

For the potassium metal, we have:

[tex]n_k = \frac{1.41g}{39g/mole}[/tex] ---where 39 is the atomic weight of potassium

[tex]n_k = 0.036\ moles[/tex]

For the liquid iodine monochloride, we start by calculating its mass using:

[tex]Mass = Density \times Volume[/tex]

So, we have:

[tex]Mass = 3.24 \times 6.52[/tex]

[tex]Mass = 21.12g[/tex]

The number of moles is then calculated as:

[tex]n_I=\frac{21.12g}{162.35}[/tex]

[tex]n_I = 0.130\ moles[/tex]

The reaction equation 2K(s) + ICl(l) → KCl(s) + KI(s) means that:

2 moles of potassium reacts with 1 mole of liquid iodine monochloride.

So, 0.036 moles of potassium will react with the following moles of liquid iodine monochloride.

[tex]A = \frac{0.036}{2}[/tex]

[tex]A = 0.018\ moles[/tex]

i.e. 0.036 moles of potassium will react with of liquid iodine monochloride

By comparison: 0.018 moles is less than 0.036 moles

So, the amount of heat liberated at constant pressure is:

[tex]Amount = 0.036 \times -\frac{740.71}{2} kJ[/tex]

[tex]Amount = -13.33 kJ[/tex]

Hence, the amount of heat liberated at constant pressure is -13.33 kJ