Question

A 90.0-mL sample of 1.00 M NaOH is mixed with 45.0

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Answer #1

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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