Question

a student mixed 25.0cm3 of 4.00moldm-3 hydrochloric acid with an equal volume of 4 dm-3 sodium hydroxide the initial temperature of both solutions was 15.0°C the maximum temperature recorded was 30.0°C​

Answer 1

Answer:

Calculate heat absorbed by the water:

(Assume density of solutions = 1.00 g/cm^3 and specific heat of solution is the same as water)

q = m c (T2-T1)

q = 50 g (4.184 J/gC) ( 30-15 C) = 3138 J

Heat released by neutralization reaction = -3138 J

Moles HCl used = 0.0250 L X 4.00 mol/L = 0.100 mol

Delta H = -3138 J / 0.100 mol = -3.14X10^4 J = -31.4 kJ/mol