Suppose 100 grams (100ml) of HCl solution were mixed with 100 grams (100 ml) of

Question

Suppose 100 grams (100ml) of HCl solution were mixed with 100 grams (100 ml) of NaoH solution (1.0 M) so that the reaction of HCl + NaOH---> NaCl + H2O occured. The calorimeter and contents had a specific heat of 1090 J/g. Suppose that the temperature of the solutions changed from 25.00 degrees celcius to 28.45 degrees celcius during the mixing.

a. calculate the moles of HCl and mole of NaOH reacting

b. The KJ of heat produced by the reaction.

C. Delta H for the reaction

d. was the reaction exothermic or endothermic?

Explanation / Answer

a) moles of HCl = 100 / 36.5 = 2.74

    moles of NaOH = 100 / 40 = 2.5

limiting reagent is NaOH.

so mole of HCl and moles of NaOH reacting = 2.5

b)   heat produced :

Q = m Cp dT + m x Cp

Q = 200 x 4.184 x (28.45-25) + 200 x 1090

Q = 221 kJ

C )

delat H = - Q / n

              = -221 / 2.5

            = -88.4 kJ / mol

d)

it is exothermic reaction

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