When 4.98 g of NaOH was dissolved in 52.79 g of water in a calorimeter at 23.7 oC, the temperature of the solution went up to 50.1 oC. What is the enthalpy change in kJ/mole of sodium hydroxide? Assume the specific heat of the mixture is the same as water.

Change in enthalpy is -51.072 kJ/mol.

Explanation:

Molar mass of NaOH = 39.997 g/mol

So, 4.98 g of NaOH = moles of NaOH = 0.125 moles of NaOH

Total mass of solution = (4.98+52.79) g = 57.77 g

Heat consumed by solution = , where m is mass C is specific heat and is change in temperature.

So, heat consumed by solution =

                                                    = 6384 J

It is an exothermic process as temperature increases during dissolution. Hence change in enthalpy should be negative.

Change in enthalpy = -(heat consumed by solution)/(no. of moles of NaOH)

                                 =

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g