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)·
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
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