I understand the first part but keep getting the second part wrong (the kJ/mol)

Question

I understand the first part but keep getting the second part wrong (the kJ/mol) (first part is in kJ/C)

Be sure to answer all parts. You want to determine anº for the reaction Zn(s) + 2HCl(aq) ZnClz(aq) + H28) To do so, you first determine the heat capacity of a calorimeter using the following reaction, whose AH is known: NaOH(aq) + HCl(aq) NaCl(aq) + H,0(1) AH° =-57.32 kJ (a) Calculate the heat capacity of the calorimeter from these data: Amounts used: 50.0 ml of 2.00 MHCl and 50.0 ml of 2,00 M NaOH Initial T of both solutions: 16.9°C Maximum T recorded during reaction: 30.4°C Density of resulting NaCl solution: 1.04 g/mL c of 1.00 M NaCl(aq): 3.93 J/gK 0.0159 Oc

Explanation / Answer

Heat released by reaction = Heat absorbed by solution(q(w)) + Heat absorbed by calorimeter(q(c))

Tempereture raise, T

T= 21.8 - 16.8 = 5

q(c) = 5°C ×(0.0159kJ/)=0.0795kJ

q(w) = (mass of HCl solution + Zn)×T× c of ZnCl2 solution

= 102.81g × 5 × 3.95J/g

= 2.03kJ

Now, calculate moles of ZnCl2 produced

Zn(s) + 2HCl(aq) ------> ZnCl2(aq) + H2(g)

Stoichiometrically, 2mole of HCl react with 1mol of Zn to produce 1mol of ZnCl2

Mass of Zn= 1.3078g

Molar mass of Zn = 65.38g/mol

moles of Zn = 1.3078g/65.38g/mol = 0.02000mol

moles of HCl = (1mol/1000ml)×100ml = 0.1mol

So, Zn is limiting reagent

stoichiometrically, 1mole of Zn produce 1mole of ZnCl2

0.02mole of Zn produce 0.0200mol of ZnCl2

Therefore,

For 1mole of ZnCl2 , Heat absorbed,q

q = (q(c) +q(w)/0.02mol)×1mol

= ((0.0795kJ + 2.03kJ)/0.02mol)×1mol

= 105.48kJ/mol

Heat absorbed = Heat released by reaction

q= -H

H°rxn= - 105.48kJ/mol

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