posted by .

Naoh (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
Using the dT determined in part 1, calculate the heat capacity of the calorimeter. Enthalpy of reaction= -58.3 kj/mol, Cs (NaCl = 3.91j/g C)density of 1 M NaCl = 1.037g/mL. dT= 12.88 degrees Celcius.50 mL of 2.00 M HCl and 50 ml of 2.05 M NaOH were used

  • Chemistry -

    What's Cs NaCl = 3.91 J/g mean?
    Is dT delta T.

  • Chemistry -

    yes dt is delta T and i think it is the specific heat capacity of NaCl but im not sure

  • Chemistry -

    You have 100 mL solution.
    Use density to calculate mass from mass = volume x density.

    mols NaCl formed = M x L = 0.2M x 0.050 = 0.1 mol. delta H rxn = 58.3 kJ/mol or 5.83 kJ for this reaction which is q in the following..
    q = [mass NaCl soln x specific heat of NaCl soln x dT] + [Ccal x dT] = 0
    Substitute and solve for Ccal.

  • Chemistry -

    Originally the enthalpy is negative then how does it become postive and I have followed the steps and I came up 47.36 J/g degrees C. Also I don't understand why enthalpy is q if you could please explain that, and thankyou so much.

  • Chemistry -

    I had to let that equation equal something and I didn't want to type that line again so I let it equal q. The usual expression for specific heat and dT calculations is
    mass H2O x specific heat H2O x (Tfinal-Tinitial) = 0. We plug in the numbers and solve for Tinitial or Tfinal or specific heat or mass H2O. In this case, however, we have all of those values and the equation equals q or any other letter I choose. I changed the sign because we are adding heat to the solution, not taking heat out; + signs mean we add heat. Another way of looking at it is that the - sign means the neutralization is an exothermic one and that it is being used to release heat and that ADDS heat to the solution.
    I went through the calculations in a hurry and came up with 47.17 J/g for Ccal I would round that to 47.2.

  • Chemistry -

    Thankyou so much for all the help and the explanation and your time.

  • Chemistry -

    I have one more question, why is the volume of NaCl 0.05 L when we are calculating the mols and 0.1 L when we are calculating mass?

  • Chemistry -

    The volume of NaCl is never 0.100.
    The mass of the SOLUTION (it is a solution of NaCl) comes from 50 mL + 50 mL = 100 mL, then
    mass of solution = volume x density = 100 mL x 1.037 g/mL = about 104 or so grams for the salt solution.

    The volume of NaCl is never 0.05 L.
    The reaction is HCl + NaOH ==> NaCl + H2O
    You have mols HCl = M x L = 2.00M x 0.050 L = 0.100 mol NaCl formed from this and the HCl is the limiting reagent.
    For the NaOH we have 2.05M x 0.050 = 0.102 mols. Of course only 0.100 mol NaCl is formed.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. chemistry

    Could you help me get a definition for each of these words. for i can have a better understanding of them Thanx Titration gas collection crystalization fractional distillation solvent limiting reagent I have defined titration, gas …
  2. Chemistry

    1.9 mol HCl and 3.8 mol NaOH react according to the equation HCl + NaOH −→ NaCl + H2O . Calculate the amount in moles of NaCl formed. Answer in units of mol
  3. Chemistry

    Help with lab report Need Help finishing my lab report.KNOWN VALUES ARE V OF HCL=50ML,M=2.188.V OF NAOH=55.3ML,M=2.088.INTIAL TEMP FOR HCL=22.5C,NAOH IS 23C.this is for part A: Mass of final NACL solution assuming that the density …
  4. Chemistry

    Naoh (aq) + HCl (aq) --> NaCl (aq) + H2O (l) Using the dT determined in part 1, calculate the heat capacity of the calorimeter. Enthalpy of reaction= -58.3 kj/mol, Cs (NaCl = 3.91j/g C)density of 1 M NaCl = 1.037g/mL. dT= 12.88 …
  5. chemistry

    When 50.0mL of 1.20 M of HCl (aq)is combined with 50.0mL of 1.30 M of NaOH (aq) in a coffee-cup calorimeter, the temperature of the solution increases by 8.01 Degrees C. What is the change in enthalpy for this balanced reaction?
  6. Chemistry

    HCl(aq) +NaOH(aq) ---> NaCl +H2O In a coffee cup calorimeter 50 ml of 2.00M HCl was combined with 50ml of 2.05M NaOH. a) What is the enthalpy change for the reaction?
  7. Chemistry

    I need help I can't seem to get the right answer. Please help! 1/2 H_2 (g) + 1/2 Cl_2 (g) --> HCl (g) -92.3 kJ/mol Na(s) + 1/2 O_2(g) + 1/2H_2(g) --> NaOH(s) -426.8 kJ/mol NaCl(s) --> Na(s) + 1/2Cl_2(g) 411.1 kJ/mol H_2O(l) …
  8. Chemistry

    Calculate the enthalpy change of formation for the following reactions using Hess's law, and the enthalpies of formation given in Figure 4: b) MgO +2HCl --> MgCl2 +H2O c)NaOH+ HCl -->NaCl +H20 Figure4: the given enthalpies of …
  9. chemistry

    In a coffee cup calorimeter, 50 mL of 0.1 M HCl and 50 mL of 0.1 M NaOH are mixed to yield the following reaction: HCl + NaOH  NaCl + H2O. The two solutions were initially at 22.1 oC and the final temperature is 24.2 oC. Calculate …
  10. chemistry

    Calculate the molar enthalpy change for this reaction: HCl(aq 1.00M) + NaOH -> NaCl(aq,.500M)+ H2O Initial temp: 22.15 degrees Celsius Extrapolated temp: 25.87 degrees Celsius DT: 3.72 degrees Celsius Notes: Calculate the enthalpy …

More Similar Questions