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Chemistry problem?
What is the temperature change when 50.0Ml of 0.500M HCL is mixed with 75.0 Ml of 0.400 M NaOH. The standard enthalpy is -57.92 Kj/mol. 4.184JGC specific heat of water is used in this with a pressure calorimeter of 22C.
1 Answer
- hcbiochemLv 71 month agoFavorite Answer
First, be careful with upper and lower case letters in chemistry. Conventions are important for communication. So, for example, I am guessing that the volume you mean is mL, hydrochloric acid is HCl, kilojoules is kJ, and the specific heat of water has units here of J/gC.
moles HCl = 0.025 mol
moles NaOH = 0.030 mol
So, HCl is the limiting reactant.
Heat released by the reaction = 0.025 mol X -57.92 kJ/mol = -1.448 kJ = -1448 J
Heat absorbed by the solution = +1448 J
q = m c (T2-T1)
1448 J = 125 g (4.184 J/gC) (T2-22 C)
T2 = 24.8 C
