Question

In: Chemistry

Find heat absorbed by water (kJ) and ΔrH (kJ mol-1) of the following reaction Mg + 2 HCl (aq) MgCl2 + H2

Find heat absorbed by water (kJ) and ΔrH (kJ mol-1) of the following reaction

Mg + 2 HCl (aq) MgCl₂ + H₂

Mass of Mg: 20.8mg

Moles of Mg: calculate using formula m/M

Ti= 19.9 C

Tf= 33.4 C

density of water: 1g/mol

5M of 1.0 M HCl

V of water is 5g

Solutions

Expert Solution

Given reaction is Mg + 2 HCl (aq) MgCl₂ + H₂

The change in enthalpy H = ?

q_(solution) = q_(reaction) = m₍_solution) C_(solution)T

m_solution = mass of the solution = 5 g

C_Solution = Specific heat capacity of water = 4.2 J/g ^oC

T = T_final - T_Initial = 33.4 - 19.9 = 13.5 ^oC

q_(reaction) = m₍_solution) C_(solution)T

= (5 g) x (4.2 J/g ^oC) x (13.5 ^oC)

= 283.5 J/g

= 0.283 KJ/g

No of moles of Magnesium = mass/Molar mass

Mass of Magnesium = 20.8 mg = 0.0208 g

Molar mass of Magnesium = 24.31 g/mol

moles of Mg = 0.0208 g / (24.31 g/mol) = 0.0008556 moles

The heat evolved for this reaction per mole of Magnesium metal = (Heat evolved in this reaction) / (Moles of Magnesium metal)

= 0.283 KJ / (0.0008556 moles) (Note it is the heat per 24.31 g magnesium)

= 330.762 KJ/mole

The heat energy produced in this reaction per mole of Magnesium metal = 330.762 KJ/mol

queen_honey_blossom answered 1 year ago

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