Part a)

Reaction of Metallic magnesium with acid is an exothermic reaction

Part b)

We have, Moles of \(\displaystyle{M}{g}={\frac{{\text{Mass}}}{{\text{Molar mass}}}}={\frac{{{0.524}}}{{{24.305}}}}={0.02156}\ {m}{o}{l}\)

Now, mass of the solution = Mass of water + Mass of Mg = 0.524 + 60.0 = 60.524 g

\(\displaystyle{q}=-{m}{c}\triangle{T}\)

Where q is the heat released

c is the specific heat

\(\displaystyle\triangle{T}\) is the temperature change

Therefore, \(\displaystyle{q}={60.524}\times{4.184}\times{\left({65.8}-{22.0}\right)}=-{11091.6}\ {J}=-{11.09}\ {K}{J}\)

Part c)

Heat released per mole of \(\displaystyle{M}{g}={\left({\frac{{-{11091.6}}}{{{0.02156}}}}\right)}=-{514}{K}\frac{{J}}{{m}}{o}{l}\)