A 105.2 mL sample of 1.00 M NaOH is mixed with 52.6 mL of 1.00 M H2SO4 in a larg

Question

A 105.2 mL sample of 1.00 M NaOH is mixed with 52.6 mL of 1.00 M H2SO4 in a large Styrofoam coffee cup; the cup is fitted with a lid through which passes a calibrated thermometer. The temperature of each solution before mixing is 21.65 °C. After adding the NaOH solution to the coffee cup and stirring the mixed solutions with the thermometer, the maximum temperature measured is 32.30 °C, Assume that the density of the mixed solutions is 1.00 g/mL, that the specific heat of the mixed solutions is 4.18J/gC), and that no heat is lost to the surroundings. 1st attempt Part 1 (1 point) Inl See Periodic Table Write a balanced chemical equation for the reaction that takes place in the Styrofoam cup. Remember to include phases in the balanced chemical equation. Part 2 (1 point) Is any NaOH or H2SO4 left in the Styrofoam cup when the reaction is over? Choose one O A. Yes B. No Part 3 (1 point) ? See Hint Calculate the enthalpy change per mole of H2SO4 in the reaction. kJ/mol

Explanation / Answer

2NaOH(aq) + H2SO4(aq) ------------> Na2So4(aq) + 2H2O(aq)
no of moles of NaOH = molarity * volume in L
                      = 1*0.1052   = 0.1052moles

no of moles of H2So4   = molarity *volume in L
                       = 1*0.0526 = 0.0526 moles
total volume of solution = 105.2 + 52.6 = 157.8ml
density of solution = 1g/ml
mass of the solution   = volume * density
                       = 157.8*1 = 157.8g
q = mcDT
    = 157.8*4.18*(32.3-21.65)
    = 7024.78J
    = 7.02478Kj

NaOH : H2O4 = 2:1   ( moles form balanced equation)
             = 0.1052: 0.0526   (moles
NaoH and H2So4 moles are sonsumed> NO left
B. NO
part-3
     DH = 7.02478KJ/0.0526moles   = 133.55KJ/mole >>>>answer

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