What is the heat change in enthalpy for the reaction (kJ/mol) of Mg(OH)2?

Question

A common antacid called "Milk of Magnesia" uses magnesium hydroxide to neutralize excess hydrochloric acid in the stomach. The name comes from the fact that magnesium hydroxide has low solubility in water and thus is administered as a suspension (milky appearance). When you combine a suspension of .729 g Mg(OH)₂ (58.33 g/mol) in 25.0 mL of water with 50.0 mL of 1.00 HCL in a constant pressure calorimeter, the temperature rises by 8.81 C. What is the heat change in enthalpy for the reaction (kJ/mol) of Mg(OH)₂? (assume no heat is lost, and the specific heat capacity of the solution is 4.184 J/g x C, and the density of the solution is 1.00 g/mL).

1. I'm a little confused on this problem, but would I use the equation where H_rxn = q = -q_soln with q = mc(Tf_f-T_i)?

Answer 1

No, you don't need to use q = mC(Tf-Ti), since you don't know either temperature. You do, however know ∆T which is 8.81º. So, you use q = mC∆T

Mg(OH)₂ + 2HCl ==> MgCl₂ + 2H₂O .. balanced equation

moles Mg(OH)2 = 0.729 g x 1 mol / 58.33 g = 0.01250 moles

moles HCl = 50.00 ml x 1 L / 1000 ml x 1.00 mole/L = 0.0500 moles HCl

Since the mole ratio of Mg(OH)2 to HCl is 1 : 2, we see that Mg(OH)2 is limiting, as it will only neutralize 0.025 moles of HCl. We will need to use this fact when we calculate the ∆Hrxn on a per mole basis.

q = mC∆T

q = heat = ?

m = mass = 25.0 ml + 50.0 ml = 75.0 ml x 1 g / ml = 75.0 g (not including the mass of Mg(OH)2 & HCl)

C = specific heat = 4.184 J/gº

∆T = change in temperature = 8.81º

q = (75.0 g)(4.184 J/gº)(8.81º)

q = -2765 J .. this is the heat liberated (- sign) by reaction of 0.0125 moles Mg(OH)₂

To get ∆H_rxn per mole, we simply divide by moles of Mg(OH)₂...

∆H_rxn = -2765 J / 0.0125 moles

∆H_rxn = -221,200 J/mol

∆H_rxn = 221 kJ/mol (3 sig. figs.)