Paul is conducting an experiment where he has combined 45.0 grams of CaO with 10
ID: 894327 • Letter: P
Question
Paul is conducting an experiment where he has combined 45.0 grams of CaO with 100.0 mL of water at 25°C,?
releasing some amount of steam as well. This produces the following equation:
CaO(s)+H20(l)-->Ca(OH)2(s)
a.) Which of the two reactants are in excess and how many grams of this reactant will remain after the reaction is complete?
b.) Using the information provided in the table below:
Substance Delta(Hf)(KJ/mol)
CaO(s) -635.1
H2O(l) -285.84
Ca(OH)2(s) -986.09
Assuming all heat is transferred to the water, what is the mass of steam that was released?
Explanation / Answer
m = 45g CaO
MW CaO = 56.0774 g/gmol
V = 100 ml
D = 1 g/ml
find moles of each substance
N CAO = mass/MW = 45/56.0774 = 0.802 mol of CaO
N H2O = V*D /MW = 100 ml* 1g/ml / 18g G/mol = 5.556 mol of CaO
Ratio of reaciton is 1:1 so CaO is limiting
CaO(s)+H20(l)-->Ca(OH)2(s)
b)
Find the total heat
H = Hprod -Hreact
H = -986.09 - ( -635.1 -285.84) = -65.15 kJ/rxn
ASsume
-Qlost = Qwin
Q = n*Cp*dT
Find excess water
5.556- 0.802 = 4.754 gmol of Water are left or 4.75*18 = 85.57 grams
This will happen:
- Q will heat water to 100°C
- Q left will evaporate water at 100°C
Heat the water
Q = m*Cp*dT
Q = (85.57 g)*4.18 J/gC * (100-25) = 26826.2 J or 26.83 kJ
The total heat is given as
Q = H*n = (-65.15 kJ/rxn)*(0.802 mol reacted) = -522.5 kJ
Since 26.873 kJ will be used to heat all the water mass to 100°C, take it away
522.5 -26.83 = 495.67 kJ are left
Now lets see how much can we boil with that
Q= m*LH
LH = latent heat of water = 2260 kJ/kg
495.67 kJ = m*2260 kJ/kg
m = 495/2260 = 0.2193 kg
or 219.3 grams may be boiled
Since we have 85.57 g of water, all water will boil