10 mol/h of 50 mol % NaOH is mixed with enough 6.5 mol % NaOH to produce a 10 mol % aqueous NaOH solution. The NaOH and water are initially at 25 OC. Assume the heat capacity of solution is 2.8 J/g C. How much of heat to be removed to keep the final solution at 10OC?

105.4 kJ

Explanation:

Given that:

10 mol/h of 50 mol % NaOH is mixed with enough 6.5 mol % NaOH to produce a 10 mol % aqueous NaOH solution

Now,if the moles of 6.5 mol% of NaOH is mixed with 10 mol of 50% NaOH

We have; Total moles = (M + 10) moles

moles of NaOH = ( 10 × 0.5) + ( M × 0.065)

The mixture then tends to be the the mole of NaOH solution

=

=

The total moles = 10 +

=

=

10% mole of NaOH & 90% moles of H₂O

Thus,    

=

=  g of NaOH

For H₂O

=

=

= g of H₂O

Finally, Total mass of mixture =

=

Heat capacity of solution = 2.8 J/g ° C

At 25 ° C, heat removed to change the temperature from 25° C to 10° C

ΔT =  ( 25 - 10 ) ° C

ΔT = 15 ° C

ΔQ = mcΔT

ΔQ = * 2.8 J/g ° C * 15° C

ΔQ = 105444 J

ΔQ = 105.4 kJ

Hence, the amount of heat to  be removed to keep the final solution at 10 ° C = 105.4 kJ

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890