molar heat of neutralization?

Question

The molar heat of neutralization of a strong acid and a strong base is determined by mixing 50.0 mL of 1.0 M HCl with 50.0 mL of 1.1 M NaOH in a coffee cup calorimeter. The initial temperatures of two solutions was 25.32°C and the final temperature inside the calorimeter was 31.43°C. What is the molar heat of neutralization? Assume that the specific heat capacity of the solution is identical to water and the density is 1.00 g/mL.

Answer 1

From the reaction you had heat produced

Q=m*Cp*ΔT

for water Cp= 1 calorie/(gram*Celsius)

You have a total of 50+50 =100mL and since d=1.00 g/mL you have 100 g

so Q =100*1*(31.43-25.32) = 611 cal

you mixed mole HCl= C*V= 1*0.05= 0.05 with
mole NaOH =C*V =1.1*0.05 =0.055 which means that, since the stoichiometry of the reaction is 1:1, HCl is the limiting reagent

0.05 mole HCl give 611 cal
1 mole HCl gives x
x=611/0.05= 12220 cal/mole =12.22 Kcal/mole

Answer 2

Heat Of Neutralization

Answer 3

This Site Might Help You.

RE:
molar heat of neutralization?
The molar heat of neutralization of a strong acid and a strong base is determined by mixing 50.0 mL of 1.0 M HCl with 50.0 mL of 1.1 M NaOH in a coffee cup calorimeter. The initial temperatures of two solutions was 25.32°C and the final temperature inside the calorimeter was 31.43°C. What is the...

Answer 4

For the best answers, search on this site https://shorturl.im/XvfeV

moles NaOH = M x L moles HCl = M x L Determine limiting reagent (in this case whichever one has fewer moles) Heat released (q) = [moles of limiting reagent)] x [-55.84kJ/mol] = __kJ Your answer should be negative because it is an exothermic reaction, meaning heat is released from the system. Good luck on your Thermodynamics lab ;)

Answer 5

rude