[hunter457]

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).


(a) I figured out that the equation is Mg_(s) + 2H₂O_(l)  Mg(OH)₂ + H_2(g) and that it is a redox reaction (hydrogen is reduced, magnesium is oxidized).

(b) How do I go about calculating this?

[Borek]

How much water was consumed by the reaction?

And read answer here:

http://www.chemicalforums.com/index.php?topic=36688.0

[hunter457]

(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?***

[Borek]

You are correctly approaching part of the problem, but for some unknown reason you decided to ignore half of my post.

How much water was consumed by the reaction?

[hunter457]

Wasn't 0.19mL consumed by the reaction? I'm confused.

[Borek]

No.

Try to solve other question first:

What mass of water reacts with 0.0272g of Mg? What is volume of this water?

[hunter457]

0.0272g / 24.31g/mol=1.11888 x 10^-3 mol Mg
(1.11888 x 10^-3)(2) = 2.23776 x 10^-3 mol H₂O
(2.23776 x 10^-3 mol)(18.016g/mol)= 0.0403g H₂O reacts with 0.0272g Mg

V=m/d=(0.0403g)/(1.00g/mL)= 0.0403mL of H₂O

So water consumed=0.0403mL and water produced=0.19mL. So then my ∆V=0.19mL-0.0403=0.1497mL

Do I then plug this into the equation where i previously had 0.19mL? What do I do after this to solve for the enthalpy change?

[Borek]

Do I then plug this into the equation where i previously had 0.19mL?

Yes.

Later you should pray that your teacher have not overlooked fact that reaction uses some of the water

What do I do after this to solve for the enthalpy change?

What is the heat of fusion of ice?

[hunter457]

"What is the heat of fusion of ice?"

It's 6.01kJ/mol

m= -0.1497mL/(-0.0905)=1.654g of ice melted

Now that I have these two values, what equation do i use?  :delta:HvesselA = - :delta:HvesselB ? How would I use this?

[Borek]

How much heat was consumed by melting ice?

Where did this heat came from?

[hunter457]

(1.654g) / (18.016g/mol) = 0.0918mol
0.0918mol x 6.01kJ/mol = 0.55176kJ of heat consumed. Is that right? Is that the enthalpy?

"Where did this heat come from?"

Did it come from vesselA and transfer to vesselB because the reaction is exothermic?

[Borek]

You don't have two vessels. If anything you have two processes - one produces energy, other consumes it, but it all happens in one calorimeter.

I have not checked numbers but what you did seems OK so far - yes, that is amount of heat that was consumed by the meting of the ice.

What is the process that produced the heat?

[hunter457]

My answer was 0.55176kJ, but the question says that it should be in kJ/mol.

"What is the process that produced the heat?"

Isn't the process the exothermic reaction of magnesium with excess water?

[Borek]

My answer was 0.55176kJ, but the question says that it should be in kJ/mol.

Please... Junior is 25 now, I don't remember how to spoon feed. How many moles reacted?

"What is the process that produced the heat?"

Isn't the process the exothermic reaction of magnesium with excess water?

Yes, no vesselA nor vesselB. Could be that's a way you were doing similar questions, but you shouldn't use such approach blindly.

[hunter457]

Ok, so the answer is  0.55176kJ/2mol =0.27588kJ/mol

Thanks so much for all of your *delete me*