Strontium oxide reacts with liquid water to produce strontium hydroxide. SrO (s)
ID: 917901 • Letter: S
Question
Strontium oxide reacts with liquid water to produce strontium hydroxide.
SrO (s) + H2O (l) Sr(OH)2 (aq)
When 2.18 g of SrO is added to 375.4 mL of water, the temperature of the solution rose from 24.87°C to 26.68°C in a coffee cup calorimeter. Assuming the density of water at this temperature is 1.000 g/mL, the specific heat of the solution is 3.949 J/(ºC•g) and the heat absorbed by the constant pressure calorimeter is negligible, calculate the molar enthalpy (in kJ/mol) for this reaction.
Please show your work. In order for me to understand. Thanks!
Explanation / Answer
Given the mass of SrO = 2.18 g
Molecular mass of SrO = 103.6 g/mol
Hence moles of SrO = mass / molecular mass= 2.18 g / 103.6 g/mol = 0.02104 mol SrO
Since the temperature of water is increased hence heat is released during the dissolution of SrO(s).
Mass of water = Vlume x density = 375.4 mL x 1.000 g/mL = 375.4 g
specific heat of the solution, s = 3.949 J/(ºC•g)
Now the heat absorbed by water can be calculated as, Q = mxsxdT
= 375.4 g x 3.949 J/(ºC•g) x(26.68 - 24.87) DegC
= 2683 J
Hence heat evolved when 0.02104 mol SrO(s) is dissolved = 2683 J
Now the heat evolved when 1 mole of SRO(s) is dissolved = (2683 J / 0.02104 mol SrO) x 1 mol SrO
= 127531 J/mol = 127.53 KJ/mol (answer)
Hence molar enthalpy is 127.53 KJ/mol