Slake lime is produced by treating quick lime with water .

Question

CaO(s) + H20(s) = Ca(HO)2(s) -65.2kj
Quick lime is a white solid that melts at 2570 degrees Celsius,. The exothermic reaction of quick with water and the rather small specific heats of both quicklime (0.946j/g.°c) and slake lime (1.20j/g.°c) made the storage and transport of the substance harazou in the old days.
A 500g sample of water is reacted with an equimolar amount of the product, Ca(OH)2? Assum that the product absorbs all the heat released in the reaction.

Answer 1

The amount of heat released in the reaction can be calculated using the equation:

q = m * c * ΔT

Where:
q = heat released (in joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)

First, let's calculate the heat released in the reaction between 500g of water and an equimolar amount of Ca(OH)2:

1. Calculate the number of moles of water:
Molar mass of H2O = 18g/mol
Number of moles of H2O = mass/molar mass = 500g / 18g/mol = 27.78 mol

2. Since the reaction is stoichiometrically balanced and the reaction is exothermic, the amount of heat released will be the same as the heat absorbed by the product. Thus, we can calculate the heat absorbed by Ca(OH)2 using the enthalpy change provided:

q = -65.2 kJ

3. Now, let's calculate the final temperature reached by the product:

q = m * c * ΔT
-65.2 kJ = 500g * 1.20 J/g°C * ΔT
ΔT = -65.2 kJ / (500g * 1.20 J/g°C)
ΔT ≈ -108.67 °C

Since temperature cannot be negative, we can conclude that the final temperature of the product will be 0 °C after absorbing all the heat released in the reaction.