Question

initial temperature = 24.3 degree celcuis

final temperature = 35.7 degree celcuis

50.0 ml of 0.740 M phospheric acid +50.0ml of 2.00M of sodium hydroxide

change in temperature = 11.4 degree celcius

calculate the following

1- what is the mass mixture in thecup ?

2-heat flowof the system in cup in KJ?

3- heat of system in cup (qrxn)in KJ?

4-initial moles of H3PO4 in the cup?

5- initial moles of NaOH in the cup?

6-delta H rxn in KJ(experimental)?

7-delta Hrxn in KJ (theoretical)

8- percent error?

please i need to help with the steps not just the results

thank you

Answer 1

1)

mass of mixture in the cup = 100 g

2)

heat Q = m Cp dT = 100 x 4.184 x (35.7 - 24.3)

heat = 4.76 kJ

4)

initial moles of H3PO4 = 50 x 0.740 / 1000 = 0.037

5)

initial moles of NaOH = 50 x 2.00 / 1000 = 0.5

6)

H3PO4 + 3 NaOH   ------------------> Na3PO4 + 3 H2O

    1              3           

0.037          0.5

here limiting reagent is H3PO4 .

delta H rxn = - Q / n = - 4.76 / 0.037 =

                 = - 128.6 kJ