Grade 12 chemistry
posted by tessa on .
This is the second reaction:
HCl(aq)+NaOH(aq)>NaCl(aq)+H2O(l)
(Heat of neutralization)
This reaction involves mixing two solutions: 1.00 mol/L NaOH and 1.00 mol/L HCl.
Trial 1: 48.0 mL of the NaOH solution is mixed with 47.5 mL of the HCl. The temperature rises from 22.00°C to 27.90°C.
HINT:
Remember that you are heating water. Use the formula Q = C x M x DT. The volume of water used equals the volume of whatever solution you are using, be it NaOH or HCl. The volume must be converted to mass (1 mL of water has a mass of 1 g). The mass must be reported in kilograms.
In the second reaction you are using a solution of NaOH. To calculate the number of moles you must use the equation c= n/V. Also the final mass of water is the sum of the volumes of the NaOH solution and the HCl solution.
Questions:
1) Calculate the heat involved in the trial(Q)
2) Calculate the moles of NaOH used in the trial(mol)
3)Calculate how much heat would be evolved if 1 mole of NaOH was used (Q/mol)

Just follow the instructions given in questions 1, 2, and 3 along with the hint which tells you what to do.
Post your work if you have trouble. 
I don't understand. in the formula Q=C*M*deltaT, do you use 4.200 kj/mol as C. Is the answer to question one, calculate the heat involved in the trial, 1.19 KJ?

The heat capacity of water is 4.18 J/g*K which is 4.18 kJ/kg*K (which I assume is why the question asks for mass in kg.
You have 48.0 mL + 47.5 mL = 95.5 mL and since the density of water is 1 g/mL, the solution has a mass of 95.5 g or 0.0955 kg.