Enthalpy of reaction between magnesium and water in ice calorimeter

[hunter457]

Homework Statement

In an ice calorimeter, the reaction between 0.0272g of magnesium with excess water was carried out. The volume of water in the calorimeter decreased by 0.19mL during the reaction.
(a) Write the equation for the reaction which occurs. Is it a redox reaction?
(b) Calculate the enthalpy of the reaction (in KJ/mol).

Homework Equations

q=mc∆t

The Attempt at a Solution

(a) I figured out that the equation is Mg(s) + 2H2O(l) Mg(OH)2 + H2(g) and that it is a redox reaction (hydrogen is reduced, magnesium is oxidized).

(b) ∆V=V2-V1=Volume of water produced by melting-volume of ice melted

mass of ice melted=mass of water produced= m

At 0degrees celsius, density of water=1g/mL, density of ice=0.917g/mL

∆V=(mass of water produced/density of water) - (mass of ice melter/density of ice)
= m(1/1 - 1/0.917)
= m x (-0.0905)
m= ∆V/-0.0905= -0.19/(-0.0905)= 2.099g of ice melted

∆Hsystem=0
∆Hsystem=∆HvesselA+∆HvesselB=0
∆HvesselA = -∆HvesselB

***Is this right so far? Where do I go from here?***

 

[lifesci2010]

I think you need to find the enthalpy for melting ice into water: 6.01 kJ. Then take the mass you calculated, divide by the molar mass of water to get moles of water, and then multiply the moles by 6.01 kJ to get the heat of water (dH = q/n). Then take the q and divide by the moles of magnesium to find the heat of reaction. I think that's how you do it.