How much heat, in kilojoules, is associated with the production of 273kg of slaked lime, Ca(OH)2?

CaO(s)+H2O(l)-->Ca(OH)2(s)
delta H=-65.2kJ

so what u do is u use pv + nrt to find number of moles then u use q = mc delta t to find the amount of heat. And that should give u the answer. (Don't doubt my knowledge as I am a former member of the 2012 US IChO Team).

So you get 65.2 kJ heat from 1 mole (74.09 g) Ca(OH)2.

-65.2 kJ x (273/74.09) = ??

What he said, except you actually multiply by positive 65.2kJ.

In order to solve this, you need to relate the ratio of kJ of heat and the moles of Ca(OH)2 in the reaction, to an unknown kJ of heat and moles of Ca(OH)2 in 273kg. 1 mole of Ca(OH)2 = 74.09g

Set up a proportion:

65.2kJ/74.09g = ??kJ/273000g

Then cross multiply the 65.2 and the 273000, and divide by the 74.09

Well, I guess making lime isn't as "lime-y" as it seems! So, according to the reaction you provided, the molar ratio between Ca(OH)2 and the heat produced is 1:1. In other words, for every 1 mole of Ca(OH)2 produced, we have -65.2 kJ of heat released.

To find the amount of heat associated with the production of 273 kg of slaked lime, we need to convert the mass to moles first. The molar mass of Ca(OH)2 is approximately 74.1 g/mol.

Since 1 kg = 1000 g, the number of moles (n) of Ca(OH)2 can be calculated as follows:

n = mass / molar mass
n = 273000 g / 74.1 g/mol

Now that we have the number of moles, we can calculate the heat released using the molar ratio:

Heat = n * delta H
Heat = (273000 g / 74.1 g/mol) * (-65.2 kJ/mol)

Now you just have to crunch the numbers, and hopefully, the answer won't make you "burn" out!

To calculate the amount of heat associated with the production of 273 kg of slaked lime, Ca(OH)2, we need to use the molar mass and stoichiometry of the reaction.

First, let's calculate the number of moles of Ca(OH)2 in 273 kg.
The molar mass of Ca(OH)2 = 40.08 g/mol (for CaO) + 2 * 16.00 g/mol (for H2O) = 74.08 g/mol.

Converting 273 kg to grams, we get:
273 kg * 1000 g/kg = 273,000 g.

Now, divide the mass (in grams) by the molar mass to find the number of moles:
273,000 g / 74.08 g/mol = 3,683.20 mol

According to the balanced equation, the reaction produces Ca(OH)2 with a heat change of -65.2 kJ per 1 mol of Ca(OH)2.

Therefore, to calculate the total heat change, we multiply the number of moles by the heat change per mole:
3,683.20 mol * (-65.2 kJ/mol) = -240,234.24 kJ

So, the production of 273 kg of slaked lime, Ca(OH)2, is associated with a heat change of approximately -240,234.24 kJ.