Paul is conducting an experiment where he has combined 45.0 grams of CaO with 10
ID: 894537 • 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
water density = 1 g /ml
water volume = 100 ml
water mass = density x volume = 1 x 100 = 100g
CaO + H2O --------------------> Ca(OH)2
56g 18 g
45g 100g
for 100 g water = 56 x 100 /18 = 311 g CaO needed but we have 45 g of CaO . so limiting reagent is CaO
water reacted = 45 x 18 / 56 = 14.46 g
remaining mass of H2O = 100 -14.46 = 85.54
85.54 g H2O remians after reaction
(b)
heat released
delta H = product detl H - reactant delta H
= -986.09 - (-635.1 - 285.84)
= -65.15 kJ
-65.15 kJ /mol heat released .