How do I solve for the delta H\'s in the boxes? I\'ve been trying to figure them

Question

How do I solve for the delta H's in the boxes? I've been trying to figure them out for a few hours now but I can't seem to get the correct answers

ave occurred if the mixina albrimeter contente e UUm is constant and lina- s the extrannlatio Chem. 1E Thermochemistry Experiment 6 Data and Results: Name: sec: Lab Partner: Data LoggerD Turn this and the foltowing pages In only with your plots. This tab is due the following week In fab Part A Reaction 1: Heat of Solution KNO, Initial Temperature AT (from plot) Mass KNO3. Temperature Probe ID: 20.I Final Temperature 1:17 -not from the dot above) 10675 H20 volume: 760 t±0.5 mL) Mass of solution (g): (show all calculations) 7067775e H experimental: (show all calculations) CAT 1820 6076 07 kI/mol AH (theoretical): (see page 8) (show all calculations) KJ/moi Percent Error: (show all calculations) Page 11 of 15 Rev. F17

Explanation / Answer

Htheoretical needs to be looked up from page 8

Calculate % error = (Htheoretical - Hcalculated)/ Htheoretical * 100

a.

H of HCl and NaOH solution:

q = m Cp T

= 166 * 4.184 * 6

= 4167.26 J

Total moles reacted = Moles of H+ (limiting reagent) = 1.63

Hexp = 4167.26 / 1.63 J/mol

= 2.56 kJ/mol

b.

For MgO formation, need T from plot values to calculate H value

c.

Mg (s) + 2 HCl (aq) MgCl2 (aq) + H2 (g); H°1 = -394kJ /mol

MgO(s)+2HCl(aq)MgCl2(aq)+H2O(l); H°2 = - 51 kJ /mol

H2 & 1/2 O2 --> H2O, H°3 = -285.9 kJ /mol

To find Hf0 (MgO):

Mg(s) + ½ O2 --> MgO (s)

From Hess’s Law:

Hf0= H°1 ­­- H°2 + H°3

= -394 kJ/mol + 51 kJ/mol - 285.9 kJ/mol

= -628.9 kJ/mol

Hf0theoretical = -602 kJ/mol

% error = (-602 + 628.9)/602 * 100

= 4.47 %

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