What are the steps to this basic thermodynamics equation?

Question

What are the steps to solving this problem?
Honestly I'm not looking for the answer, I want to actually know the steps on how to solve this sort of problem. Thanks :)

A 0.582-g sample of Mg reacts with excess 1.0 M HCl (60.0 mL) according to the procedure used in this experiment. The initial and final temperatures were 24°C and 68°C. What is the ∆H of the reaction per mole of magnesium? Assume that the specific heat capacity for the solution is 4.187 J/deg∙g and that the density of the 1.0 M HCl is 1.00 g/mL. (Hints: the magnesium contributes to the mass of the solution)

Cameron Andrews · Accepted Answer

This is how I would proceed: Write down the actual reaction. You should anticipate a single replacement reaction when magnesium, a monatomic substance, reacts with hydrochloric acid, a diatomic substance: #color(white)("Mg"(s) + 2"HCl"(aq) -> "MgCl"_2(aq) + "H"_2(g))#
(highlight when you figure it out) 2) Examine your available variables and determine which equation they fit into. You are given what you need to determine #DeltaT# (#"44 K"# or #""^@ "C"#), the mass of the liquid that the metal is immersed into, and you are given the specific heat capacity. That ought to bring to mind: #\mathbf(q = mcDeltaT)# where 3) Using the information you were given, determine how to find these variables. 4) Find the relationship between enthalpy #H# and heat flow #q#. We are under constant pressure, as you can now see, so this relationship is valid: #\mathbf(DeltaH = q_p)# where #DeltaH# is the change in enthalpy and #q = q_p# at a constant pressure. From this, since we want the enthalpy per #"mol"#, let us divide both sides by the number of #"mol"#s, #n#: #\mathbf(DeltabarH) = (DeltaH)/n = \mathbf(q_p/n)# where #barH# is the molar enthalpy and #n# is the number of #"mol"#s. (All we did was divide by #n#.) Furthermore, we should determine what #n# is for. We want to divide by the number of #"mol"#s, but of what?  Of whatever reactant gives us the maximum yield *possible. That should be the *limiting reagent (magnesium, of course, since #"HCl"# is in excess).  Therefore, to calculate the molar enthalpy: #color(blue)(DeltabarH = q_p/("mols of limiting reagent"))# 5) Work out your concluding equation: #color(white)(DeltabarH = ((m_"Mg" + m_"soln")cDeltaT)/("mols of limiting reagent"))# 
(highlight when you figure it out) In the end, #q# should be #color(white)"11.2" "kJ"#, and the enthalpy should be #color(white)(466)# #"kJ/mol"#.

Olivia Anton · Answer

undefined Sure, could you please provide the specific basic thermodynamics equation you're referring to?