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)

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

The given parameters are:

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

For the potassium metal, we have:

---where 39 is the atomic weight of potassium

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

So, we have:

The number of moles is then calculated as:

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.

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:

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

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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

- both b and d.

-since b and d are the closest coastlines then that makes it the most affected.

< 3

catya that's not even a b or c