5oocm3 of 1.0 mol dm3 of sodium hydroxide solution at room temperature was placed in a styrofoam cup and 25cm3 of 2.0 mol-dm3 of hydrochloric acid was added at the same temperature.The temperature rose by 15.2 degrees celsius. Calculate the heat of neurtralization
NaOH + HCl ==> NaCl + H2O
mols NaOH = M x L = 1.0 x 0.5 = 0.5
mols HCl = M x L = 0.025 x 2 = 0.05
So HCl is the limiting reagent.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Find q. That is q/0.05 mol H2O.
Convert to kJ/mol
This is the right way to do it but I don't think it gives the correct answer. Check the problem to be sure the numbers are right. Could that be 1.52 C and not 15.2?
To calculate the heat of neutralization, we need to use the formula:
Heat of Neutralization = (molarity of acid) * (volume of acid) * (change in temperature) / (number of moles of reacting acid)
First, let's determine the number of moles of hydrochloric acid used. The molarity (concentration) of the hydrochloric acid is given as 2.0 mol/dm3, and the volume used was 25 cm3. We can convert the volume to dm3 by dividing by 1000:
Volume of hydrochloric acid = 25 cm3 = 25/1000 dm3 = 0.025 dm3
Now, we can calculate the number of moles of hydrochloric acid:
Number of moles of hydrochloric acid = (molarity) * (volume)
= 2.0 mol/dm3 * 0.025 dm3
= 0.05 moles
Next, let's determine the number of moles of sodium hydroxide used. The molarity of the sodium hydroxide solution is 1.0 mol/dm3, and the initial volume used was 50 cm3. Again, we can convert the volume to dm3:
Volume of sodium hydroxide solution = 50 cm3 = 50/1000 dm3 = 0.05 dm3
Number of moles of sodium hydroxide = (molarity) * (volume)
= 1.0 mol/dm3 * 0.05 dm3
= 0.05 moles
Now, we can calculate the heat of neutralization:
Heat of Neutralization = (molarity of acid) * (volume of acid) * (change in temperature) / (number of moles of reacting acid)
= (2.0 mol/dm3) * (0.025 dm3) * (15.2 degrees Celsius) / (0.05 moles)
= 7.6 degrees Celsius
Therefore, the heat of neutralization is 7.6 degrees Celsius.