Pb(NO3)2(aq) + 2 NaBr(aq) → PbBr2(s) + 2 NaNO3(aq)
no of moles Pb(NO3)2 = molarity * volume in L
                                  Â
= 0.75*0.2 = 0.15 moles
no of moles of NaBr   = molarity * volume in
L
                                   Â
= 1.5*0.2 = = 0.3
0.15 moles of Pb(NO3)2 react with 0.3 moles of NaBr to gives
0.15 moles of PbBr2
total volume = 200+ 200 = 400ml
mass of solution = volume * density
                            Â
= 400*1 = 400g
q = mC deltaT
     = 400*4.2*2.44  =
4099.2 J
    J/mole = 4099.2/0.15  =
27328J/mole = 27.328KJ/mole
