Chemistry, 15.07.2020 17:01 triss6

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)

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Answer from: dnarioproctor

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

The given parameters are:

Mass of potassium metal = 1.41gAmount of liquid iodine monochloride = 6.52 mL

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

$n = \frac{Mass}{Atomic\ weight }$

For the potassium metal, we have:

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

$n_k = 0.036\ moles$

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

$Mass = Density \times Volume$

So, we have:

$Mass = 3.24 \times 6.52$

Mass = 21.12g

The number of moles is then calculated as:

$n_I=\frac{21.12g}{162.35}$

$n_I = 0.130\ moles$

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.

$A = \frac{0.036}{2}$

$A = 0.018\ moles$

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:

$Amount = 0.036 \times -\frac{740.71}{2} kJ$

Amount = -13.33 kJ

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

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