2. A strip of magnesium metal having a mass of 1.22 g dissolves in 100.0 mL of 6.02 M HCl
which has a density of 1.10 g/mL. The hydrochloric acid is initially at 23.0 °C and the
resulting solution reaches a final temperature of 45.5 °C. The heat capacity of the
calorimeter in which the reaction occurs is 562 J/°C. Calculate the ∆H for the reaction under
conditions of the experiment, assuming the specific heat of the final solution is the same as
that for water.

Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g)

In this question heat capacity of the calorimeter is already given.
Heat capacity = ms = C
First you have to calculate how many moles of MgCl2 is formed?
Convert g of Mg to moles of Mg to moles of MgCl2
Convert Molarity *Volume  of HCl to moles of HCl to moles of MgCl2
check which one is the limiting reactant ?
Calculate
q (heat) =ms ∆t= C. ∆t ( change it to KJ as ∆H is in KJ/mole)
You know C and  ∆t already so plug in the values in the above equation.
To know ∆H  which is KJ / mole
∆H = q / n     ( here n is the number of moles of MgCl2)

First you have to calculate how many moles of MgCl2 is formed?
Convert g of Mg to moles of Mg to moles of MgCl2
>>>>>>1.22 g Mg*(1 mol Mg/24.33)= 0.05 mol Mg
>>>>>>0.05 mol Mg*(1 mol MgCl2/ 1 mol Mg)= 0.05 mol MgCl2
Convert Molarity *Volume  of HCl to moles of HCl to moles of MgCl2
>>>>>>0.1 L * 6.02 M= .602 mol HCl
>>>>>>0.602 mol HCl *(1 mol MgCl2/ 2 mol HCl)=0.301 mol MgCl2
check which one is the limiting reactant ?
>>>>>>Mg is the limiting reactant for the reaction, so the moles of the product, MgCl2 would be 0.05 mol as well.
Calculate
q (heat) =ms ∆t= C. ∆t ( change it to KJ as ∆H is in KJ/mole)
>>>>>>>>>ms= C
>>>>>>>>> q= C * ∆t = .562 KJ/C * 22.5 C= -12.645 KJ
To know ∆H  which is KJ / mole
∆H = q / n     ( here n is the number of moles of MgCl2)
>>>>>>> q= -12.645 KJ
>>>>>>> n of MgCl2= .05 mol
>>>>>>> ∆H= -12.645 KJ/ 0.05 mol = -252.9 KJ/mol
The sign of 252.9 KJ/mol will be negative because it is exothermic as the ∆t is positive.
However, the answer in the manual for q is -23.1 KJ, and the answer for ∆H is -461 KJ/mol. Please let me know if there were any errors in my calculation process.

This is not a question of bomb calorimeter
Different parts of calorimeter (container/thermometer/stirrer /cover etc) can also absorb heat as they are not 100% insulated.
In this case total heat released
q= C ∆t(due to heat capacity of calorimeter) +  ms ∆t(rise in temperature of the solution)
here m is the total mass of the solution = 100mL 1.10g/mL +1.22g
C ∆t = 562*22.5J
s is 4.18J/g degC
now get the value of q
you will get 23.10 kJ
convert it in ∆H