a)
the balanced reaction is given by
H2S04 (aq) + 2 NaOH (aq) ---> Na2S04 (aq) + 2H20 (l)
b)
we know that
moles = molarity x volume (L)
so
moles of NaOH taken = 1 x 90 x 10-3 = 90 x 10-3
moles of H2S04 taken = 1 x 45 x 10-3 = 45 x 10-3
now
consider the reaction
H2S04 (aq) + 2 NaOH (aq) ---> Na2S04 (aq) + 2H20
(l)
we can see that
moles of NaOH reacted = 2 x moles of H2So4 taken
so
moles of NaOH reacted = 2 x 45 x 10-3 = 90 x 10-3
so
all the NaOH and H2S04 is reacted
so
No , H2S04 or NaOH is left
c)
total volume = 90 + 45 = 135
now
mass = density x volume
so
mass of solution = 135 x 1 = 135 g
now
heat = m x s x dT
so
Q = 135 x 4.18 x ( 30.5 - 23.7)
Q = 3837.24 J
now
Qrxn = - Q = -3837.24
now
enthalpy change = -Q / moles of H2S04
enthalpy change = -3837.24 / 45 x 10-3
enthalpy change = -85.272 x 1000
enthaly change = -85.272 kJ /mol
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