all parts for all points Heat of Neutralization for a strong acid-strong base re

ID: 793848 • Letter: A

Question

all parts for all points

Heat of Neutralization for a strong acid-strong base reaction

1.mass of NaOH solution: 51.651 g

2. mass of HCl solution: 50 g

3. mass of neutralized solution: 101.604 g

4. change in temp of solution: 6.8 degrees celsius

Find

a.heat gained by the combined solutions

b.heat released in the reaction

c.molarity of NaOH

d.molarity of HCl solution

e. volume of NaOH solution used in the reaction

f. volume of HCl solution used in the reaction

g.moles of OH- involved in n=m*v

h.moles of H+ involved in the reaction

i. moles of water produced in the reaction

j.heat released in the reaction per mole of water produced

k.averge change in H(n)

a) heat of strong acid(HCl) - strong base neutralisation = -57 KJ/mol

NaOH mole s= 51.651/40 = 1.29

HCl moles = 50/36.5 = 1.37

since HCl and NaOH react in 1:1 ratio i.e HCl + NaOH <-> NaCl + H2O

moles neutralised =1.29 , heat gained by combined soln = 57 x1.29 = 73.53 KJ

b) heat released = -73.53 KJ ( -ve sign indicates heat released)

c) NaOH molarity = moles/total vol = ( 1.29 x1000/66.6) = 19.37 ( total vol calculations shown in next steps )

d) HCl Molarity = 1.37 x1000/66.6 = 20.57

e) NaOH vol used = mass/density = 51.651/2.13 = 24.25 ml

f) HCl vol used = 50/1.18 = 42.37 ml

( total vol = 24.25+42.37 = 66.6 ml)

g) OH- moles = 19.37 x 66.6 /1000 = 1.29

h) H+ moles = 20.57 x 66.6/1000 = 1.37

i) water moles = 1.29

heat absorbed by solution = mass x temp change x spcific heat = 101.65 x 6.8 x 4.184 = 2891 KJ= 2.891 KJ

j) heat released per 1 mole water = (73.53-2.89)1.29 = 54.76 KJ/mole

k) average change in dH = - 73.53+2.89 = -70.64 KJ

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