Question

Consider the following reaction:

2 HCl(aq) + 1 Mg(s) → 1 MgCl₂(aq) + 1 H₂(g) ΔH°_rx = -465.8 kJ

Determine the amount of heat released, in kJ, of each of the following scenarios. Give each answer to 2 decimal places.

(a) 10.69 g of Mg(s) is reacted with excess HCl(aq)·
(c) 1.17 g of H₂(g) is produced from the reaction after completion.
(b) Excess Mg(s) is added to 440 mL of 0.996 M of HCl(aq)·
(d) 580 mL of a 0.729 M solution of MgCl₂(aq) is produced from the reaction after completion.
(e) 10.21 g of Mg(s) is added to 175 mL of 0.687 M of HCl(aq)·

Answer 1

2HCl(aq) + 1Mg(s) -----> 1MgCl2(aq) + 1H2(g) ∆H°_rxn = -465.8kJ

a)

Stoichiometrically, reaction of 1mole of Mg release 465.8kJ of heat

givene moles of Mg = 10.69g/ 24.305g/mol = 0.4398mol

Heate released by the reaction of 0.4398mol of Mg =

(465.8kJ/1mol) × 0.4398mol

= 204.9 kJ

c)

Stoichiometrically, 1mole of H2 production release 465.8kJ of heat

number of moles of H2 = 1.17g/ 2.016g/mol = 0.5804mol

heat released by production of 0.5804moles of H2 =0.5804mol × 465.8kJ/mol

= 270.4 kJ

b)

Stoichiometrically, reaction of 2moles of HCl releases 465.8kJ of heat

moles of HCl = ( 0.996mol/ 1000ml) × 440ml = 0.4382mol

Heat released by the reaction of 0.0004382moles of HCl =

465.8kJ/2mol × 0.4382mol

= 102.1 kJ

d)

Stoichiometrically, this production of 1mole of MgCl2 release 465.8 kJ of heat

moles of MgCl2 produced = ( 0.729mol/1000ml) × 580ml = 0.4228mol

heat released by 0.4228moles of MgCl2 = 465.8kJ/mol × 0.4228mol

= 196.9 kJ

e)

Number of moles of Mg = 0.4201mol

Number of moles of HCl = ( 0.687mol/1000ml) × 175ml = 0.1202mol

limiting reagent is HCl

stoichiometrically, reaction of 2moles of HCl release 465.8kJ of heat

heat released by 0.1202moles of HCl = (465.8kJ/2mol)× 0.1202mol

= 27.99 kJ